C程序中的分段错误（核心转储）错误

Question

I tried to compile and run the following program to reverse a string using the gcc compiler for linux but it shows the error : segmentation fault (core dumped).I even tried to debug using gdb but it didn't help. The program given below firstly inputs t which is the number of test cases.I tested the program with 3 test cases but after taking the 2nd input from user, the compiler shows error.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* strrev(char*);

int main(int argc, char *argv[])
{
        int t,i=0,temp=0;
        char *str[10],*rev[10];
        scanf("%d",&t); //input the number of test cases
        while(i<t)
        {
            scanf("%s",str[i]);
            i++;
        }
        while(temp<t) //reverse the string and display it
        {
           rev[temp]=strrev(str[temp]);
           printf("%s \n",rev[temp]);
           temp++;
        }
    return 0;
    getchar();
}

Function to reverse the string:

char *strrev(char *str) 
{

    int i = strlen(str)-1,j=0;

    char ch;
    while(i>j)
    {
        ch = str[i];
        str[i]= str[j];
        str[j] = ch;
        i--;
        j++;
    }
    return str;
} 

Answer 1

You are getting segmentation fault because you haven't allocated space for elements of str . You need to allocate memory first in main function.

scanf("%d",&t); //input the number of test cases

if(t <= 10)
    for(size_t i = 0; i < t; i++)
        str[i] = malloc(50); // Assuming string is no more than 50characters. 
else  
    exit(0);  

Beside this there are many flaws in your code. Here is the code after fixing them

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void strrev(char*);  // Change return type to void

int main(void)
{
    int t,i=0,temp=0, ch;
    char *str[10];
    scanf("%d",&t); //input the number of test cases
    while((ch = getchar()) != EOF && ch != '\n'); // To consume newline character after scanf

    // Allocate memory for str elements
    if(t <= 10)
        for(size_t i = 0; i < t; i++)
            str[i] = malloc(50); // Assuming string is no more than 50characters.
    else
        exit(0);

    i = 0;
    while(i < t)
    {
        fgets(str[i],50,stdin); // Use fgets instead of scanf to read string
        i++;
    }
    while(temp<t) //reverse the string and display it
    {
        // Since you are reversing string by flipping the characters the same 
        // string just pass pointer to it. str[temp] will be updated in function.
        strrev(str[temp]);
        printf("Reverse is %s \n", str[temp]);
        temp++;
    }
    return 0;
}

void strrev(char *str)
{

    size_t i = strlen(str)-1,j=0;
    char ch;
    while(i>j)
    {
        ch = str[i];
        str[i]= str[j];
        str[j] = ch;
        i--;
        j++;
    }
    //printf("Reverse is %s \n", str);
}

Answer 2

you have missed to allocate memory before reading char* value, so you can do this:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* strrev(char*);

int main(int argc, char *argv[])
{
        int t,i=0,temp=0;
        char *str[10],*rev[10];
        scanf("%d",&t); //input the number of test cases
        while(i<t)
        {
            str[i] = (char*)malloc(100); // just allocate memory
            scanf("%s", str[i]);
            i++;
        }
        while(temp<t) //reverse the string and display it
        {
           rev[temp]=strrev(str[temp]);
           printf("%s \n",rev[temp]);
           temp++;
        }
    return 0;
    getchar();
}

Answer 3

char *str[10],*rev[10];

You did not assign storage to hold string values yet for those pointers.

char * str; /* this is a string pointer */

char * str = malloc(15); /* this creates storage for a string */

char str[10]; /* this creates a static char array, also is a string */

Answer 4

I also had the same issue. I just fixed it by correcting the indices of the matrix.