why i can't print this array with using pointers?

i am trying to enter string values (more than 1 for each array) for a1 and a2. I want to print them row by row but the program don’t print any of the values. How can i fix it?

#include<stdio.h>

int main (void){
int i;
char *a1[10], *a2[10];


for (i=0; i<10;i++){
    printf("Enter text %d for a1", i);              
    scanf("%s",a1);
    printf("Enter text %d for a2", i);
    scanf("%s",a2);
}

for (i=0; i<10;i++){
                    
    printf("text %d is %s for a1",i,a1[i]);
    printf("text %d is %s for a2",i,a2[i]);
}

return 0;


}

i tried to print elements of arrray row by row but i cant. Also the program doesn’t give an error. Program ending after entering the elements.

>Solution :

1. Your program doesn’t compile as you never defined mecut.
2. Use constants instead of hard coding magic values.
3. a1 and a2 are array of pointers. You need to allocate space for the strings you want to read in.
4. The two scanf() overwrite data into the same variable but you want it be relative to the index.
5. When reading a string with scanf() ensure you set the maximum string length to avoid buffer overflow. Consider using fgets().
6. Check the return value of scanf() otherwise your variables might not be initialized.
7. For readability use \n in your printf().

Here is working code:

#include <stdio.h>
#define N 10
#define STR_LEN 99
#define STR(s) STR2(s)
#define STR2(s) #s

int main (void){
    char a1[N][STR_LEN+1];
    char a2[N][STR_LEN+1];
    for (int i=0; i<N; i++){
        printf("Enter text %d for a1: ", i);
        if(scanf("%" STR(STR_LEN) "s", &a1[i]) != 1) {
            printf("scanf failed\n");
            return 1;
        }
        printf("Enter text %d for a1: ", i);
        if(scanf("%" STR(STR_LEN) "s", &a2[i]) != 1) {
            printf("scanf failed\n");
            return 1;
        }
    }
    for (int i=0; i<N; i++) {
        printf("text %d is %s for a1\n",i, a1[i]);
        printf("text %d is %s for a2\n",i, a2[i]);
    }
}

and example run:

$ seq 20 | ./a.out 
Enter text 0 for a1: Enter text 0 for a2: Enter text 1 for a1: Enter text 1 for a2: Enter text 2 for a1: Enter text 2 for a2: Enter text 3 for a1: Enter text 3 for a2: Enter text 4 for a1: Enter text 4 for a2: Enter text 5 for a1: Enter text 5 for a2: Enter text 6 for a1: Enter text 6 for a2: Enter text 7 for a1: Enter text 7 for a2: Enter text 8 for a1: Enter text 8 for a2: Enter text 9 for a1: Enter text 9 for a2: text 0 is 1 for a1
text 0 is 2 for a2
text 1 is 3 for a1
text 1 is 4 for a2
text 2 is 5 for a1
text 2 is 6 for a2
text 3 is 7 for a1
text 3 is 8 for a2
text 4 is 9 for a1
text 4 is 10 for a2
text 5 is 11 for a1
text 5 is 12 for a2
text 6 is 13 for a1
text 6 is 14 for a2
text 7 is 15 for a1
text 7 is 16 for a2
text 8 is 17 for a1
text 8 is 18 for a2
text 9 is 19 for a1
text 9 is 20 for a2

Here is a refactored version that uses a couple of functions to reduce duplication. Also using fgets() instead of scanf() as you may not like how the latter by default reads words opposed to lines:

#include <stdio.h>
#include <string.h>
#define N 10
#define STR_LEN 100

int prompt_str(int index, const char *name, size_t len, char s[len]) {
    printf("Enter text %d for %s: ", index, name);
    if(!fgets(s, len, stdin)) {;
        return 1;
    };
    s[strcspn(s, "\n")] = '\0';
    return 0;
}

void print_str(int index, const char *name, const char *s) {
    printf("text %d is %s for %s\n", index, s, name);
}

int main (void){
    char a1[N][STR_LEN];
    char a2[N][STR_LEN];
    for (int i=0; i<N; i++){
        if(
            prompt_str(i, "a1", STR_LEN, a1[i]) ||
            prompt_str(i, "a2", STR_LEN, a2[i])
        )
            return 1;
    }
    for (int i=0; i<N; i++) {
        print_str(i, "a1", a1[i]);
        print_str(i, "a2", a2[i]);
    }
}