Thread: Defining Integers Etc.

Why are you declaring all variables as type int?
One should be type int, one should be type float and the other one should be type char.

There is no need to use flushall. Remove that line.

Code:

  #include <stdio.h>

int main(void)
{
    /* The three variables we are going to identify*/

     int myint;
     int mychar[2];
     float myfloat ;

    /*Identifies a number*/

    printf("Enter a number: ");
    scanf("%d", &myint);

    /*Identifies a character*/

    printf("Enter a character: ");
    getchar();
    scanf("%c", &mychar);

    /* Identifies a float*/

    printf("Enter a float: ");
    scanf("%f", &myfloat);

    printf("%d === %c === %f\n",myint,mychar,myfloat);
}

The problem happening while reading the char .Because usually scanf will read the input upto '\n'.so when you read the integer the '\n' will remain in the buffer.When you try to read a character that '\n' comes in to play .To avoid that we can use getchar() to take that '\n' character . then you can read the character using scanf .

There is no standard flushall() function, but guessing by its name, it is probably useful here. The problem is that after entering an integer, the user hits enter. The scanf("%c") will grab the newline, not allowing the user to enter a character (this is the source of the common-to-new-programmers problem of “my getchar() is being skipped”). flushall() presumably will “clear” standard input, removing the newline.

Such a call isn't needed before %f because %f will eat leading whitespace.

printf("Enter a number: ");
scanf("%d", &myint);

You're not checking if scanf was succesfull.

What if i enter "bad code" and then enter? myint wont receive a number, character will be 'b' and float will fail aswell.

Originally Posted by karthigayan

Code:

  #include <stdio.h>

int main(void)
{
    /* The three variables we are going to identify*/

     int myint;
     int mychar[2];
     float myfloat ;

    /*Identifies a number*/

    printf("Enter a number: ");
    scanf("%d", &myint);

    /*Identifies a character*/

    printf("Enter a character: ");
    getchar();
    scanf("%c", &mychar);

    /* Identifies a float*/

    printf("Enter a float: ");
    scanf("%f", &myfloat);

    printf("%d === %c === %f\n",myint,mychar,myfloat);
}

The problem happening while reading the char .Because usually scanf will read the input upto '\n'.so when you read the integer the '\n' will remain in the buffer.When you try to read a character that '\n' comes in to play .To avoid that we can use getchar() to take that '\n' character . then you can read the character using scanf .

Code:

printf("Enter a character: ");
    getchar();
    scanf("%c", &mychar);

This will result in compiling error. You're passing a pointer to pointer to scanf.

Thanks for the help!

Well i did well on that but when i print out the variables i get a different character then i input. This is the code i used. Weirdd

Code:

  #include <stdio.h>

int main(void)
{
    /* The three variables we are going to identify*/

     int myint;
     int mychar[2];
     float myfloat ;

    /*Identifies a number*/

    printf("Enter a number: ");
    scanf("%d", &myint);

    /*Identifies a character*/

    printf("Enter a character: ");
    getchar();
    scanf("%c", &mychar);

    /* Identifies a float*/

    printf("Enter a float: ");
    scanf("%f", &myfloat);

    printf("myint = %d\n",myint );
    printf("mychar = %c\n",mychar) ;
    printf("myfloat = %f",myfloat);
    
    getch();
}

Originally Posted by grumpy

I think it's weird that you haven't fixed the problems pointed out to you, but you expect the code to work anyway.

getchar() discards a character. scanf("%c", &mychar) reads a subsequent character from the input. mychar also needs to be of type char, not an array of two integers.

So should i remove the pointer getchar()? I was told to put that in instead of flushing the previous input..

Originally Posted by grumpy

I see such things fairly often with novices who incorrectly believe a pointer and an array are the same thing.

But, his assignment never even asked for a pointer or an array so it baffles the mind why he is defining one.