how to replace a macro with a string?

Question

#include <stdio.h>
#include <stdarg.h>
#define ammo "full"

int main()
{
  char a[100];
  a = ammo;
  printf("%s",a);
  return 0;
}

I am trying to replace #define ammo with "full" (string) and want to print it on the screen to check if it works but the code is not compiling.

Now, [pick up a good book](https://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list). This is among the basic things one learns when starting. — StoryTeller - Unslander Monica, Aug 28 '17 at 12:30
the macro is replaced by a string by the _pre-processor_. What you are actually want is to assign that string to `a` — CIsForCookies, Aug 28 '17 at 12:31
`*a=ammo` does not work. What Filip mentioned was `char *a = ammo;` which is totally different as it is definition with initialization instead of an assignment. — Gerhardh, Aug 28 '17 at 12:33
@karan - No, `*a=ammo` doesn't work because `a` is an array and not a pointer. *Please* read a book, it works much faster than trial-and-error. — Bo Persson, Aug 28 '17 at 12:33
@karanshaarma Because a string is essentially a **sequence of characters**. A pointer can represent it by pointing to the 1st character. — iBug, Aug 28 '17 at 12:40
For all the bitter SO veterans who've been reading this without casting close votes, check out the C FAQ below the [C tag wiki](https://stackoverflow.com/tags/c/info). It contains many useful canonical duplicates, including "How do I copy strings?" and "How do I compare strings?". — Lundin, Aug 28 '17 at 15:03

Math Lover · Answer 1 · 2017-08-28T12:35:19.160

2

Instead of a=ammo; you should use strcpy(a,ammo); you should not use a direct assignment when you have strings, but use the C method strcpy to copy a string to another. it will work

edited Aug 28 '17 at 12:35

answered Aug 28 '17 at 12:32

Math Lover

148
10

For a better answer, I think you should add the 2 other options provided by StoryTeller and Filip Kočica :) – CIsForCookies Aug 28 '17 at 12:34

iBug · Answer 2 · 2017-08-28T12:42:32.547

It's invalid to assign a C-string to an array (except when initializing, see below). You must "copy" the string to the array because the space is already allocated:

strcpy(a, ammo);

Better yet, use a safer version of copy function:

strncpy(a, ammo, sizeof(a) / sizeof(char));

Or directly assign it when initializing:

char a[100] = ammo;

Or, don't use array. Use a pointer instead:

char *a;
a = ammo;

Note you can't change the content of the string if you use a pointer.

how to replace a macro with a string?

2 Answers2