CODEWITHSUNDEEP
c
M3Lee OP
M3Lee OP

Reputation: 33

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

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" 
}

Upvotes: 0

Views: 83

Answers (2)

Nick S
Nick S

Reputation: 1369

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.

Upvotes: 1

4386427
4386427

Reputation: 44274

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;
}

Upvotes: 2

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