将字符串数组作为参数传递给 Function

Question

I am trying to pass an array of strings to a function and then the print it there. But it is giving segmentation fault.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
    char *str[2];
    for(int i=0;i<2;i++) {
            scanf("%s",(str+i));
    }
    display(str);
}

void display(char **p)
{
    for(int i=0;i<2;i++) {
            printf("%s \n",p[i]);
    }
}

Answer 1

I think the problem is that you are not allocating memory for your list. Here is a sample code that i made.

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

#define LIST_SIZE 2
#define MAX_WORD 100

void display(char**);

int main() {
    char buffer[100];
    char** list;
    int i;
    int n;

    // allocate
    list = (char**)malloc(sizeof(char*) * LIST_SIZE);

    for(i = 0; i < LIST_SIZE; i++) {
        scanf("%99s",buffer);
        n = strlen(buffer);
        list[i] = (char*)malloc(sizeof(char) * n);
        strcpy(list[i], buffer);
    }

    display(list);

    return 0;
}

void display(char** str) {
    int i;
    printf("-- output ---\n");
    for(i = 0; i < LIST_SIZE; i++) {
        printf("%s\n", str[i]);
    }
}

I made the list only allocate the space necessary for the word. If you want fixed you dont need a buffer and the strcpy.

Answer 2

For starters, you need to make the following change in the "display" function....

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

In addition, the "display" function needs to be placed above "main" or you need to declare a prototype for the function.

Answer 3

What about this:

void display(char * p[])

This way you can pass the argument