scanf trouble: easy code, inexperienced programer

When ever I use scanf("%d", &n) ; and then have a following printf("Generic Statement:"); My compiler displays the printf statement without me inputting a value for n.

Code:

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int x, y;main()
{printf("\nInput an integer value for x:"); 
  scanf("%d",  &x); 

printf("\nInput an integer value for y:"); 
scanf(" %d",  &y);       
      if (x==y)
        printf("x is equal to y"); 
      if (x>y) 
        printf("x is greater than y");         
      if (x<y)
        printf("x is less than y");
 
 return 0;
}

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This is an example from my book, and it doesn't work.....how frustrating
Such that before I input a value, this is displayed as the output

Code:

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Input an integer value for x:
Input an integer value for y:
x is equal to y

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Why doesn't the complier wait for the scanf user-input before finishing the code?

Quote:

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When ever I use scanf("%d",n) ; and then have a following printf("Generic Statement: %d", n); My compiler displays the printf statement without me inputting a value for n.

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That's strange. Compilers usually translate code from one language to another.

Quote:

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This is an example from my book, and it doesn't work.....how frustrating

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You should get a better book like K&R2. Unless you understand buffering, it's not a good idea to try to synchronise data on two 'interactive' streams.

The book is OK. Your keyboarding has left you adrift.

Code:

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//When ever I use
scanf("%d",n) ;  //you forgot the ampersand: scanf("%d",&n);

//and then have a following
printf("Generic Statement: %d", n); //you forgot the name of the variable: printf"Generic Statement: %d\n",INT VARIABLE NAME HERE);

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It's a good habit to put a \n on the very end of all printf() statements - some systems have a buffer for it, and it will drive a programmer bonkers!

Quote:

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Originally Posted by Adak

The book is OK.

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Err, no. Why do x and y need static storage duration other than to stop your program from breaking when scanf fails?

Quote:

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It's a good habit to put a \n on the very end of all printf() statements - some systems have a buffer for it, and it will drive a programmer bonkers!

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I'm not sure what you mean by "some systems have a buffer for it". His code doesn't take into consideration that stdout may be line buffered when connected to an interactive device, and that can be fixed either by writing a new line to stdout or calling fflush.

I forgot to include that ampersand in the forum text.... whoops

Quote:

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Originally Posted by Adak

The book is OK. Your keyboarding has left you adrift.

Code:

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//When ever I use
scanf("%d",n) ;  //you forgot the ampersand: scanf("%d",&n);

//and then have a following
printf("Generic Statement: %d", n); //you forgot the name of the variable: printf"Generic Statement: %d\n",INT VARIABLE NAME HERE);

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It's a good habit to put a \n on the very end of all printf() statements - some systems have a buffer for it, and it will drive a programmer bonkers!

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One caveat to this: When I enter a value for x and y respectively; the correct printf statement will appear. However, this is only true after I see the odd premature printf display as noted in the original post

Quote:

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Originally Posted by ProBallChamp

When ever I use scanf("%d",n) ; and then have a following printf("Generic Statement: %d", &n); My compiler displays the printf statement without me inputting a value for n.

Code:

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int x, y;main()
{printf("\nInput an integer value for x:"); 
  scanf("%d",  &x); 

printf("\nInput an integer value for y:"); 
scanf(" %d",  &y);       
      if (x==y)
        printf("x is equal to y"); 
      if (x>y) 
        printf("x is greater than y");         
      if (x<y)
        printf("x is less than y");
 
 return 0;
}

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This is an example from my book, and it doesn't work.....how frustrating

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There is nothing in that code which will cause a "premature printf display". Unless your expectations are wrong.

Try providing a exact description of the input you are supplying when you run the program, and the results you get.

We're not mindreaders, you know.

A couple of minor points unrelated to your problem

1) It is a good idea for there to be a #include <stdio.h> as the first line.

2) Explicitly give main an int return type. Yes, I know some books tell you the compiler assumes that anyway, but it's a bad habit for a programmer to rely on that, roughly on the level with you picking your nose and eating it.

3) The compiler does not display printf() statements. Your program does.

This comes from a review of a book, actually, which is irrelevant unless you are reading that book, but it contains the best explanation of the problem I think you are experiencing now that I have ever seen.

As you should know, C is a standardized language --

Quote:

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Schildt frequently omits trailing newlines on output. This is a common habit among DOS programmers, as the MS-DOS command prompt apparently used to start with a newline, so that programs which had already produced a newline were followed by a blank line.

However, it's not portable, in two ways. The first is that, in general, C does not define the behavior of output to a text stream which is never terminated by a newline. ("Whether the last line requires a terminating new-line character is implementation-defined."; C99, 7.19.2, paragraph 2.)

The second is that streams may be buffered -- output sent to a stream may not be delivered to the host environment until flushed (using fflush()) or until some other event occurs. There are three levels of buffering; unbuffered (self-explanatory), line-buffered (data are buffered until a new-line character is encountered), or fully buffered (data are buffered until a certain size of block is filled, such as "every 8,192 bytes".) The standard input and output streams are very often line-buffered, although the standard does not require this; all it says that standard input and standard output are "fully buffered if and only if the stream can be determined not to refer to an interactive device." (C99, 7.19.3, paragraph 7).

What this means is that, on many systems, printing a prompt which does not end with a newline produces no visible output until you either explicitly flush the stream or send something ending in a newline. (There may be other circumstances under which output is sent, but only those are guaranteed to deliver the output to the host environment.) One common exception is that some systems will automatically flush output streams when waiting for input; on such systems, the prompt would actually get displayed. This behavior is strongly recommended by C99.

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C: The Complete Nonsense

Here is as far a I know how to go (since I am new to the C programming language)

Code:

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#include <stdio.h>
#include <math.h>
int x, y;
main()
{    printf("\nInput an integer value for x:");
      scanf("%d", &x);
      printf("\nInput an integer value for y:");   
      scanf(" %d", &y);
        if (x==y)
          printf("\nx is equal to y");
        if (x>y)
          printf("\nx is greater than y");
        if (x<y)
          printf("\nx is less than y\n");
    return 0;}


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The output is as such

Code:

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Input an integer value for x:
Input an integer value for y:
x is equal to y

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How can I prevent "x is equal to y"from being displayed prior to me inputting anything?

Quote:

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Originally Posted by grumpy

There is nothing in that code which will cause a "premature printf display". Unless your expectations are wrong.

Try providing a exact description of the input you are supplying when you run the program, and the results you get.

We're not mindreaders, you know.

A couple of minor points unrelated to your problem

1) It is a good idea for there to be a #include <stdio.h> as the first line.

2) Explicitly give main an int return type. Yes, I know some books tell you the compiler assumes that anyway, but it's a bad habit for a programmer to rely on that, roughly on the level with you picking your nose and eating it.

3) The compiler does not display printf() statements. Your program does.

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As I said before .....

Quote:

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Originally Posted by grumpy

Try providing a exact description of the input you are supplying when you run the program, and the results you get.

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You have described the output repeatedly but provided absolutely NO DESCRIPTION of the INPUT you are supplying. In your program, the input affects the output.

Does it include digits? Does it include whitespace (spaces, carriage returns, etc)? Does it include other characters?


The text that whiteflags quoted is interesting, but would not explain "premature" output ..... it would explain output coming AFTER you would expect.

Try seeing how this performs:

Code:

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#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int x, y;

    puts("enter two values; the value of x, followed by the value of y");
    if (scanf("%d %d", &x, &y) != 2)
        return EXIT_FAILURE;
    printf("%d %d\n", x, y);
    return 0;
}

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edit: BTW, chances are you're entering something that can't be converted by scanf, causing it to fail on both accounts and immediately proceed through the program. Since you've statically allocated x and y, they'll both have an initial value of 0, causing them to be equal if a value doesn't get stored in them by scanf. That's usually the kind of behaviour you'd expect when you ignore its return value.

ok this is what i see.

Code:

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int x,y;

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you have it set as global, instead that should go after

Code:

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main()
{

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also

Code:

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main()

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is very bad. seeing your using the return right
change your main to

Code:

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int main(void)

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it is better practice, and some new compiliers yell at you if you dont!

also you dont need to include math.h, as your really not using anything from the math header

now in the end, even without these changes, the code runs on my pelles c ide

after those "fixes" whatever compilier your using should accept it and run it right.

if not, turn UP all the warnings and error notifications and maybe it will warn you what is wrong.

Okay, I see what you are looking for, here are three working cases when I input a x and y respectively;
1)
Input:

Code:

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5  /* x  */
4  /* y  */

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Output

Code:

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Input an integer value for x:
Input an integer value for y:
x is greater than y

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2)
Input:

Code:

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4
5

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Output

Code:

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Input an integer value for x:
Input an integer value for y:
x is less than y

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3)
Input:

Code:

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5
5

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Output

Code:

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Input an integer value for x:
Input an integer value for y:
x is equal to y

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Quote:

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Originally Posted by grumpy

As I said before .....

You have described the output repeatedly but provided absolutely NO DESCRIPTION of the INPUT you are supplying. In your program, the input affects the output.

Does it include digits? Does it include whitespace (spaces, carriage returns, etc)? Does it include other characters?


The text that whiteflags quoted is interesting, but would not explain "premature" output ..... it would explain output coming AFTER you would expect.

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So what's the problem?

Crossfire Could you clarify this for me.

Code:

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int main(void){int x, y;

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So the above is not redundant; saying integer main and integer variable x and y?

also what does the main(void) mean/do?

Quote:

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Originally Posted by Crossfire

ok this is what i see.

Code:

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int x,y;

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you have it set as global, instead that should go after

Code:

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main()
{

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also

Code:

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main()

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is very bad. seeing your using the return right
change your main to

Code:

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int main(void)

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it is better practice, and some new compiliers yell at you if you dont!

also you dont need to include math.h, as your really not using anything from the math header

now in the end, even without these changes, the code runs on my pelles c ide

after those "fixes" whatever compilier your using should accept it and run it right.

if not, turn UP all the warnings and error notifications and maybe it will warn you what is wrong.

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How do I get it to stop printing

Code:

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x is greater than y

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when I don't input anything?

Quote:

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Originally Posted by ProBallChamp

So the above is not redundant; saying integer main and integer variable x and y?

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The code you've quoted defines main with a return type of int, and it also defines two objects of the type int with block scope and automatic storage duration. int was implied in earlier standards when the return type of a function was omitted, but that was changed in the later standards.

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also what does the main(void) mean/do?

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Since it's a function definition, it's synonymous "main()". It means that the function accepts zero parameters. "void" just makes that intention clearer.

Quote:

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Originally Posted by ProBallChamp

Okay, I see what you are looking for, here are three working cases when I input a x and y respectively;

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From information you've given, none of the three cases you've supplied strike me as exhibiting a problem.

Quote:

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Originally Posted by ProBallChamp

How do I get it to stop printing

Code:

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x is greater than y

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when I don't input anything?

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It does that when you input a number greater than zero followed by end-of-file or rubbish, not when you don't input anything. You fix that by reading #11.

Quote:

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Originally Posted by ProBallChamp

How do I get it to stop printing

Code:

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x is greater than y

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when I don't input anything?

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You mean that the program prints "x is greater than y" before you have typed ANY input AT ALL to the program?

That probably means a configuration problem affecting your build process (i.e. the executable you are running was produced by the compiler/linker from a superseded version of your source code). Delete all the object files and executable files, and rebuild your complete program from source. In an IDE, do "Build Clean" followed by "Build All".

Okay, if think I've got a substantial program now. Though, I still have much to learn. Thanks for all the help Grumpy and Barney

Quote:

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Originally Posted by grumpy

You mean that the program prints "x is greater than y" before you have typed ANY input AT ALL to the program?

That probably means a configuration problem affecting your build process (i.e. the executable you are running was produced by the compiler/linker from a superseded version of your source code). Delete all the object files and executable files, and rebuild your complete program from source. In an IDE, do "Build Clean" followed by "Build All".

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so it works now?

and although i dont look for attaboys, dont forget me! :P lol

Thanks too Crossfire, thought you logged off, I now have a better understand of this concept than I did this morning.

Quote:

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Originally Posted by Crossfire

so it works now?

and although i dont look for attaboys, dont forget me! :P lol

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