如何为字符串数组创建指针？

Question

I am new to C and not sure how to pass pointers/addresses properly when it comes to chars/strings. I can't get my head around these "strings", about how to master the pointers.

I basically want to pass the address of the array "a" defined below to an arbitrary function, but I don't know how to define the pointer for this array.

Please help me!

I have the following code:

void change(char** a){
    a[0][0]='k';    //that should change a inside main
}

void main() {
    char a[2][3];
    char *tempWord;
    tempWord="sa";
    a[0]=tempWord;
    a[1]=tempWord;
    change(&a);
}

Answer 1

You have a char[2][3] so you can pass a char (*)[2][3] to your change function. To copy tempWord into your char[][] you can use strncpy . Assuming I understand what you're trying to do, that might look like

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

void change(char (*a)[2][3]) {
    *a[0][0] = 'k';
}

int main() {
    char a[2][3];
    char *tempWord = "sa";
    strncpy(a[0], tempWord, strlen(tempWord));
    strncpy(a[1], tempWord, strlen(tempWord));
    change(&a);
}

Is there any other definition of pointer other than char(*a)[2][3] ?

I guess you really did want a char ** ; you will need to malloc and free your memory for that. Something like

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

void change(char **a) {
    a[0][0]='k';
}

int main() {
    char **a;
    char *tempWord = "sa";
    a = malloc(2 * sizeof(char **));
    a[0] = malloc(strlen(tempWord) * sizeof(char *));
    a[1] = malloc(strlen(tempWord) * sizeof(char *));
    strncpy(a[0], tempWord, strlen(tempWord));
    strncpy(a[1], tempWord, strlen(tempWord));
    change(a);
    printf("%s\n", a[0]);
    free(a[1]);
    free(a[0]);
    free(a);
}