如何打印出由 C 中的 char ** 创建的字符串（数组）数组？

Question

I'm new to C and trying to create an array using char **, but I'm having a problem printing it out. Is there a way to print it out or shouldn't I use this way to create an array of string?

int main()
{
    char **a = {"abc", "ddd", "ccc", "aaa"};

    for (size_t i = 0; i < 4; i++){
        printf("%s\n", a+i);  // only print out "abc" correctly
//        printf("%s\n", *(a+i));  doesn't work
//        printf("%s\n", a[i]);  doesn't work
    }

    return 0;

}

Answer 1

This declaration

char **a = {"abc", "ddd", "ccc", "aaa"};

is incorrect. A scalar type may be initialized using only a single expression in braces.

If you want to declare an array of pointers then you should write

char * a[] = {"abc", "ddd", "ccc", "aaa"};

To output it you can write

for (size_t i = 0; i < sizeof( a ) / sizeof( *a ); i++){
    puts( a[i] );
}

Or

char * a[] = {"abc", "ddd", "ccc", "aaa"};
const size_t N = sizeof( a ) / sizeof( *a );

for (size_t i = 0; i < N; i++){
    puts( a[i] );
}

Another approach is to declare a two-dimensional array like for example

char a[][4] = {"abc", "ddd", "ccc", "aaa"};

It can be outputted the same way as shown above.

for (size_t i = 0; i < sizeof( a ) / sizeof( *a ); i++){
    puts( a[i] );
}

If you want to have a declaration of an object of the type char ** like

char **a = /*...*/;

then you could initialize it with a compound literal.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    enum {  N = 4 };
    char **a = ( char * [N] ){"abc", "ddd", "ccc", "aaa"};
    
    for ( size_t i = 0; i < N; i++ )
    {
        puts( a[i] );
    }
    
    return 0;
}

Its output is

abc
ddd
ccc
aaa

Answer 2

running the posted code through the compiler results in:

gcc -ggdb3 -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled1.c" -o "untitled1.o" (in directory: /home/richard/Documents/forum)
untitled1.c: In function ‘main’:
untitled1.c:8:17: warning: initialization of ‘char **’ from incompatible pointer type ‘char *’ [-Wincompatible-pointer-types]
    8 |     char **a = {"abc", "ddd", "ccc", "aaa"};
      |                 ^~~~~
untitled1.c:8:17: note: (near initialization for ‘a’)
untitled1.c:8:24: warning: excess elements in scalar initializer
    8 |     char **a = {"abc", "ddd", "ccc", "aaa"};
      |                        ^~~~~
untitled1.c:8:24: note: (near initialization for ‘a’)
untitled1.c:8:31: warning: excess elements in scalar initializer
    8 |     char **a = {"abc", "ddd", "ccc", "aaa"};
      |                               ^~~~~
untitled1.c:8:31: note: (near initialization for ‘a’)
untitled1.c:8:38: warning: excess elements in scalar initializer
    8 |     char **a = {"abc", "ddd", "ccc", "aaa"};
      |                                      ^~~~~
untitled1.c:8:38: note: (near initialization for ‘a’)
untitled1.c:11:18: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char **’ [-Wformat=]
   11 |         printf("%s\n", a+i);  // only print out "abc" correctly
      |                 ~^     ~~~
      |                  |      |
      |                  char * char **
Compilation finished successfully.

when compiling, do NOT ignore the warnings.

That final line from the compiler:

 Compilation finished successfully

only means the compiler used some 'workaround' for each of the problems, NOT that the correct code was produced.

If we make a change so only a single pointer is used rather than a pointer to pointer:

#include <stdio.h>




int main( void )
{
    char *a = {"abc", "ddd", "ccc", "aaa"};

    for (size_t i = 0; i < 4; i++){
        printf("%s\n", a+i);  // only print out "abc" correctly
//        printf("%s\n", *(a+i));  doesn't work
//        printf("%s\n", a[i]);  doesn't work
    }

    return 0;

}

results in:

gcc -ggdb3 -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled1.c" -o "untitled1.o" (in directory: /home/richard/Documents/forum)
untitled1.c: In function ‘main’:
untitled1.c:8:23: warning: excess elements in scalar initializer
    8 |     char *a = {"abc", "ddd", "ccc", "aaa"};
      |                       ^~~~~
untitled1.c:8:23: note: (near initialization for ‘a’)
untitled1.c:8:30: warning: excess elements in scalar initializer
    8 |     char *a = {"abc", "ddd", "ccc", "aaa"};
      |                              ^~~~~
untitled1.c:8:30: note: (near initialization for ‘a’)
untitled1.c:8:37: warning: excess elements in scalar initializer
    8 |     char *a = {"abc", "ddd", "ccc", "aaa"};
      |                                     ^~~~~
untitled1.c:8:37: note: (near initialization for ‘a’)
Compilation finished successfully.

which is better, but still not right.

modifying the declaration of a[] so it is appropriately declared and initialized results in:

char *a[]

which reads as an array of pointers to char

the next problem is:

printf("%s\n", a+i);  

which results in:

untitled1.c:17:22: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char **’ [-Wformat=]

that is corrected by:

printf("%s\n", *(a+i) );

so a passable, but not great version of the code looks like:

#include <stdio.h>




int main( void )
{
    char *a[] =
    {
        "abc", 
        "ddd",
        "ccc", 
        "aaa"
    };

    for (size_t i = 0; i < 4; i++){
        printf("%s\n", *(a+i) );  // only print out "abc" correctly
//        printf("%s\n", *(a+i));  doesn't work
//        printf("%s\n", a[i]);  doesn't work
    }

    return 0;

}

However, that has the 'magic' number 4 .

We can eliminate that 'magic' number by letting the compiler perform the calculations for the for() loop:

for( size_t i = 0; i < (sizeof(a) / sizeof( *a ); i++ )

So the final code is: (with excess blank lines removed)

#include <stdio.h>

int main( void )
{
    char *a[] =
    {
        "abc", 
        "ddd",
        "ccc", 
        "aaa"
    };

    for (size_t i = 0; i < (sizeof( a ) / sizeof( *a )); i++)
    {
        printf("%s\n", *(a+i) );  
    }

    return 0;
}

The result of running the above code is:

abc
ddd
ccc
aaa