How to pass 2d array of string to the function and print value of it?

Question

Why it is not working... It should be working, right? gcc have problem with this line, but why?

render_history(history, 2);

Sorry for bothering. I am just a beginner.

 #include <stdio.h>

void render_history(char** history, const int entry);

int main()
{
char* history[3][4];

history[0][0] = "1234";
history[1][0] = "5678";
history[2][0] = "9012";

render_history(history, 2); //??
return 0;
}


void render_history(char** history, const int entry)
{
// print "9012" 
}

Try this instead "void render_history(char history[][4], const int entry);" You have to specify the second dimension. — Austin Stephens
– Austin Stephens, Commented May 19, 2018 at 17:51
Do you know how arguments are passed to the main function? The argv argument? It's an array of strings, an array of pointers. Which seems to be what you want instead of an array of arrays of pointers. What you have now is, basically, somewhat equivalent to a 3d array. — Some programmer dude
– Some programmer dude, Commented May 19, 2018 at 17:54
Be careful, @AustinStephens: the OP has a 2D array of char * (not char), so the needed parameter type would be char * history[][4], or, equivalently, char * (*history)[4]. — John Bollinger
– John Bollinger, Commented May 19, 2018 at 18:01
As for why it doesn't work, it's because history decays to a pointer to an arrays of four pointers, char *(*)[4], which is very different from char **. If you had e.g. char history[3][5], or char *history[3] it would be a very different matter. — Some programmer dude
– Some programmer dude, Commented May 19, 2018 at 18:02

4386427 · Accepted Answer · 2018-05-19 18:30:43Z

2

gcc have problem with this line, but why?

Because the type is wrong. char* history[3][4]; can't be passed as char**. They are incompatible types.

Try something like:

#include <stdio.h>

void render_history(char* (*history)[4] , const int entry)
{
   printf("%s\n", history[entry][0]);
} 

int main()
{
    char* history[3][4];

    history[0][0] = "1234";
    history[1][0] = "5678";
    history[2][0] = "9012";

    render_history(history, 2);
    return 0;
}

edited May 19, 2018 at 18:30

answered May 19, 2018 at 18:17

4386427

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Comments

Nick S · Accepted Answer · 2018-05-19 18:52:46Z

As mentioned above double pointer not equal to 2D array. You can also use pointer to pointer of char. char **history. And with this you have several option:

1) Use compound literals

#include <stdio.h>

void render_history(const char **history, const int entry)
{
    printf("%s\n", history[entry]);
}

int main(void)
{
    const char **history = (const char *[]) { "1234", "5678", "9012", NULL};
    render_history(history, 2);  
    return 0;
}

If you need change your data later

2) Use dynamic memory allocation with malloc

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

void render_history(char **history, const int entry)
{
    printf("%s\n", history[entry]);
}

int main(void)
{
    char **history = malloc(3 * sizeof(char *));

    for (int i = 0; i < 3; ++i)
    {
        history[i] = malloc(4 * sizeof(char));
    }

    strcpy(history[0], "1234");
    strcpy(history[1], "5678");
    strcpy(history[2], "9012");
    history[3] = NULL;

    render_history(history, 2);
    return 0;
}

If you use 2nd option dont forget free memory after use.

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