segmentation fault (core dumped) error in C program - c

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.