Thread: Help needed (problem with basic C program)

you first tell us, what is your definition of "typedef"

char namearr[namespaces][names];/*declaring an array of 50 rows and each row consists of a string of 50 chars*/

array size should be a constant integer number, here "namespaces" and "names" are variable names

scanf(%s,&namearr[i][j]);

just open any book on 'c' and see syntax of scanf()

printf("The names you inputted are input[%d][%d]",i,j,namearr[i][j]);

2 access specifiers and 3 variable names ???

Originally Posted by Salem

Not really.

It was just another code dump without explanation.
Trivial to spot syntax errors mean you didn't even do the basics of making sure that it compiled.

Why is this a problem?
Because we end up spending at least a day and half a dozen messages just fixing (or guessing) whatever imaginary mistakes there are, only to be met with responses "Oh well, the real code isn't like that".

And yes, I can use google translate as well, thanks for asking.

Others can chip in, I've lost sympathy for your case.

Oh really? I am really looking for your sympathy... Please please help. Realistically I couldn't be less concerned of who you are and whether I get your dear sypathy or not, because there are alot of programmers here that are willing to help without their noses up in the sky like you.

It's a shame seeing your name was an arabic one I thought you'd understand arabic I wrote "sorry brother, I hope I haven't bothered you with my pathetic question oh polite one" but I bet google translate returned that too.

I lost all respect for you and I don't think another keystroke reply of yours will result in one keystroke of mine so save your precious time you spend replying uselessly and go learn some lessons on humbleness and how to helpful towards others. The fact that you don't know everything after C is kind of shocking, huh?

Originally Posted by gaurav9991

you first tell us, what is your definition of "typedef"



array size should be a constant integer number, here "namespaces" and "names" are variable names



just open any book on 'c' and see syntax of scanf()



2 access specifiers and 3 variable names ???

Thanks so muc for the kind reply! I see that you cannot use a variable in an array definition so do I just use namearr[namespaces[50]][names[50]];?

scanf is used as follows

scanf(%d, &int variable here);
scanf(%c, &char variable here);
scanf(%lf, &double variable here);

but I dont understand how i will incorporate the array in there. I could work my way arround before scanf by:
char inpt[50];
namearr[i][j]=inpt[50];
scanf(%s, inpt);

is that true?


and for the remaining printf i am completely clueless since the problem with array input in scanf correctly I am unsure of how I can do it with printf..

Thanks for all your help.

Originally Posted by CommonTater

Ummmm.... no that would be namearr[50][50];

The typedefs are totally nonsensical in this case.



If you are building an array of 50 strings of 50 characters each C will let you use...
scanf("%s",namearr[i]) putting it directly into your array.



Well... you're going to have to learn to look stuff up for yourself sometime...
What does the C library documentation say about printf() ?

okay but what typpe would my array be? just char namearr[50][50]?
So it would be of 50 rows each one containing a string of 50 chars?

Since C lets me use scanf("%s", namearr[i]) to input the string in the ith row into my array then so should printf when i say printf("Entry number %d is %s",i,namearr[i]); ?

printf is almost the same as scanf where to output a variable type you use %(symbol of that variable) but you can also output text.

for example i want to output a string declared as char greeting[21] and inputted my code would be

char greeting[20];
printf("please input the greeting you want me to output on the screen,-no longer than 20 characters-\n");
scanf("%s", greeting);
printf("the string you inputted is %s\n", greeting);

Thanks for the reply it's really helping

Originally Posted by EpicYuzer

okay but what typpe would my array be? just char namearr[50][50]?
So it would be of 50 rows each one containing a string of 50 chars?

Yes... that is how 2 dimensional arrays are declared. They can be any type. Remember a string in C is simply an array of characters so the result is 50 x 50 character strings.

Since C lets me use scanf("%s", namearr[i]) to input the string in the ith row into my array then so should printf when i say printf("Entry number %d is %s",i,namearr[i]); ?

It should, as long as the array bounds in i are valid.

printf is almost the same as scanf where to output a variable type you use %(symbol of that variable) but you can also output text.

for example i want to output a string declared as char greeting[21] and inputted my code would be

char greeting[20];
printf("please input the greeting you want me to output on the screen,-no longer than 20 characters-\n");
scanf("%s", greeting);
printf("the string you inputted is %s\n", greeting);

Thanks for the reply it's really helping

Ok... now it's time to move this to the next level...

1) Please learn how to use code tags when posting here... and use them.
2) Please test your code... compile it and run it before asking questions about it.

You will need these skills sometime... now is as good a time as any.

Originally Posted by CommonTater

Yes... that is how 2 dimensional arrays are declared. They can be any type. Remember a string in C is simply an array of characters so the result is 50 x 50 character strings.



It should, as long as the array bounds in i are valid.




Ok... now it's time to move this to the next level...

1) Please learn how to use code tags when posting here... and use them.
2) Please test your code... compile it and run it before asking questions about it.

You will need these skills sometime... now is as good a time as any.

Thanks so much for your feedback!

The program compiled with the following code

Code:

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

int main(int argc, char *argv[])
{
 int i, j;
  char namearr[50][50];
  
  printf("please enter the names you want in the file (limit is 50 names)");
  
  for(i=0;i<50;i++)
  {
 print
 f("Type name entry %d\n",i);
 scanf("%s",&namearr[i]);
 }
 

for(j=0;i<50;i++)
{
 printf("Input %d is %s\n",i,namearr[i]);
} 
  
  system("PAUSE");	
  return 0;
}

But it seems to always ask for strings even when I input '\0' or nothing at all. Haven't found a way around that yet.

and another when I put a for loop as follows:

Code:

for(i=0;i<50;i++)

it doesn't compile due to " 'for' loop initial declaration used outside C99 mode " error. I don't understand why...

Thanks for the help

Code:

 
for(j=0;i<50;i++)  //use i not j, 
{
 printf("Input %d is %s\n",i,namearr[i]);
}

Should help some.

Originally Posted by Adak

Code:

 
for(j=0;i<50;i++)  //use i not j, 
{
 printf("Input %d is %s\n",i,namearr[i]);
}

Should help some.

Yes thanks!


The final successful code is:

Code:

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

int main(int argc, char *argv[])
{
 int i,j;
  char namearr[50][50];
  
  printf("please enter the names you want in the file (limit is 50 names)");
  
  for(i=0;i<50;i++)
  {
 printf("Type name entry %d\n",i);
 scanf("%s",&namearr[i]);
 }
 

for(j=0;j<50;j++)
{
 printf("Input %d is %s\n",j,namearr[j]);
} 
  
  system("PAUSE");	
  return 0;
}

But still I can't find a way to stop prompting the user for strings if he has entered some number of strings less than 50 and has no more to add.

And still I don't understand why:

Code:

for(int i=0;i<50;i++)

doesnt work or compile for that matter as a for loop and returns error mentioned earlier...

Adak pointed out a nice logical error in your program.

Interestingly enough though, I didn't get such a warning when I pasted your code into a file and compiled it (gcc -Wall), but I did get the following:

test.c:14: error: format '%s' expects type 'char *', but argument 2 has type 'char (*)[50]'

You only need namearr[i], which is the address of the start of the ith element of namearray, which is a char *, which is what scanf expects.

The "'for' loop initial declaration used outside C99 mode" error usually means that you did something like

Code:

for (int i = 0; i < 50; i++)

instead of

Code:

int i;
for (i = 0; i < 50; i++)

The former is not compliant with the C standard most compilers use while the latter is legit C/C++ pretty much everywhere.

History: In 1999, they came up with the C99 standard, which allowed declaring your iterator inside the for loop, but since that standard had lots of other complicated stuff for compilers to implement (like handling imaginary numbers and type-agnostic math functions), so very few compilers ever fully implemented the standard. Thus, unless you specify something like --std=c99 (gcc), the compiler may complain about such declarations.

Originally Posted by EpicYuzer

But still I can't find a way to stop prompting the user for strings if he has entered some number of strings less than 50 and has no more to add.

Ok, here comes the "looking stuff up" part... Take a look at the stuff in your entries loop. Actually look them up in the C library documentation and study how they work... Does anything in that documentation suggest a way of ending the loop?

Code:

for(int i=0;i<50;i++)

doesnt work or compile for that matter as a for loop and returns error mentioned earlier...

That's because of the for (int i ... With the declaration inside the braces is a C-99 enhancement and you are not using a C-99 capable compiler.

Yet another case for looking stuff up.

Seriously... Do you know the 4 rules of computer bliss?

1) Read the effing screen
2) Read the effing help file
3) Read the effing manual
4) Ok, now you can ask your question.

Seriously... This has been around for years and years... Ever heard RTFM... well, that's were it comes from.