Ошибка iso c90 запрещает смешение деклараций и кода

When compiling in c, the compiler is giving me a warning telling me that C90 forbids mixing declaration and code, however their is no case of this in my code and the line that it says the warning occurs at is clearly just the declaration of a variable their is no code mixed with it.

This is what my compiler shows

gcc -c Functions.c -Wall -ansi -pedantic
Functions.c:27:18: warning: ISO C90 forbids mixing declarations and code
      [-Wdeclaration-after-statement]
        LinkedListNode* curr;
                        ^
1 warning generated.

answered Oct 25, 2017 at 21:24

I declared a variable in this way:

int i = 0;

I get the warning:

ISO C90 forbids mixed declarations and code

How can I fix it?

Up until the C99 standard, all declarations had to come before any statements in a block:

void foo()
{
  int i, j;
  double k;
  char *c;

  // code

  if (c)
  {
    int m, n;

    // more code
  }
  // etc.
}

C99 allowed for mixing declarations and statements (like C++). Many compilers still default to C89, and some compilers (such as Microsoft’s) don’t support C99 at all.

So, you will need to do the following:

1. Determine if your compiler supports C99 or later; if it does, configure it so that it’s compiling C99 instead of C89;

2. If your compiler doesn’t support C99 or later, you will either need to find a different compiler that does support it, or rewrite your code so that all declarations come before any statements within the block.

answered Nov 8, 2012 at 16:22

Just use a compiler (or provide it with the arguments it needs) such that it compiles for a more recent version of the C standard, C99 or C11. E.g for the GCC family of compilers that would be -std=c99.

To diagnose what really triggers the error, I would first try to remove = 0

• If the error is tripped, then most likely the declaration goes after the code.

• If no error, then it may be related to a C-standard enforcement/compile flags OR …something else.

In any case, declare the variable in the beginning of the current scope. You may then initialize it separately. Indeed, if this variable deserves its own scope — delimit its definition in {}.

If the OP could clarify the context, then a more directed response would follow.

-Wdeclaration-after-statement minimal reproducible example

main.c

#!/usr/bin/env bash

set -eux

cat << EOF > main.c
#include <stdio.h>

int main(void) {
    puts("hello");
    int a = 1;
    printf("%d\n", a);
    return 0;
}
EOF

Give warning:

gcc -std=c89 -Wdeclaration-after-statement -Werror main.c
gcc -std=c99 -Wdeclaration-after-statement -Werror main.c
gcc -std=c89 -pedantic -Werror main.c

Don’t give warning:

gcc -std=c89 -pedantic -Wno-declaration-after-statement -Werror main.c
gcc -std=c89 -Wno-declaration-after-statement -Werror main.c
gcc -std=c99 -pedantic -Werror main.c
gcc -std=c89 -Wall -Wextra -Werror main.c
# https://stackoverflow.com/questions/14737104/what-is-the-default-c-mode-for-the-current-gcc-especially-on-ubuntu/53063656#53063656
gcc -pedantic -Werror main.c

The warning:

main.c: In function ‘main’:
main.c:5:5: warning: ISO C90 forbids mixed declarations and code [-Wdeclaration-after-statement]
     int a = 1;
     ^~~

Tested on Ubuntu 16.04, GCC 6.4.0.

answered Oct 30, 2018 at 12:16