Looping a statically allocated array, outside of c99 mode?

Question

This is in reference to a solution posted on: Looping a fixed size array without defining its size in C

Here's my sample code:

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

int main(int argc, char *argv[])
{
    static const char *foo[] = {
           "this is a test",
           "hello world",
           "goodbye world",
           "123", 
           NULL
    };

    for (char *it = foo[0]; it != NULL; it++) {
        printf ("str %s\n", it);
    }

    return 0;

}

Trying to compile this gives:

gcc -o vararray vararray.c
vararray.c: In function ‘main’:
vararray.c:14: warning: initialization discards qualifiers from pointer target type
vararray.c:14: error: ‘for’ loop initial declaration used outside C99 mode

Answer 1

Besides the initialization in the for loop, you're incrementing in the wrong place. I think this is what you mean (note that I'm not exactly a C guru):

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

int main(int argc, char *argv[])
{
    static const char *foo[] = {
           "this is a test",
           "hello world",
           "goodbye world",
           "123", 
           NULL
    };
    const char **it;
    for (it=foo; *it != NULL; it++) {
        printf ("str %s\n", *it);
    }

    return 0;

}

Answer 2

1. Your loop variable it is of type char* , the contents of the array are of type const char* . If you change it to be also a const char* the warning should go away.

2. You declare it inside the for statement, this is not allowed in C before C99. Declare it at the beginning of main() instead.
   Alternatively you can add -std=c99 or -std=gnu99 to your gcc flags to enable the C99 language features.

Answer 3

Use -std=c99 option when compiling your code in order to use the C99 features.

Change it to const char* type ( to remove the warnings)

Answer 4

在C99之前，在for循环中声明字符指针是非标准的。

Answer 5

您需要做两件事才能在没有警告的情况下进行编译：声明iterator const char* it ，并在函数的开始而不是在loop语句中进行操作。