Why this is segmentation fault ?

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C

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

int main()
{
    char *str;
    strcpy(str,"Hello");
    printf("%s",str);
    return 0;
}

What I have tried:

I tried with the help of debugger, it shows Segmentation fault in Codeblocks IDE.
Program received signal SIGSEGV, Segmentation fault.

Posted 16-Apr-16 2:09am

Anupkumar_vj

Updated 16-Apr-16 3:31am

Dave Kreskowiak

v2

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3 solutions

Solution 3

Quote:
Why this is segmentation fault ?

By your program is done wrong.
The C language is not managed, you have to manually manage memory, and you fail to do it in your program.

C++

strcpy(str,"Hello");

Copies "Hello" to a place you don't own, that is the reason of the segment fault.

You need to learn properly the C language and follow tutos. Knowing the syntax is not enough, there is much more to know.

Posted 16-Apr-16 3:31am

Patrice T

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Jochen Arndt · Accepted Answer · 2016-04-16T02:37:00

This is because str is just an unitialised pointer to a string. There is no memory allocated where the content of the entered string can be stored and the pointer itself can point to anywhere.

So you must allocate some memory and assign that to the pointer:

C++

// It is always a good idea to initialise variables
char *str = NULL;
char *heapMemory = NULL;
char stackMemory[128];

str = stackMemory;
strcpy(str,"Hello stack");
printf("%s",str);

heapMemory = (char*)malloc(128);
str = heapMemory;
strcpy(str,"Hello heap");
printf("%s",str);
// Heap memory should be freed when not used anymore
free(heapMemory);

OriginalGriff · Accepted Answer · 2016-04-16T02:36:00

Simple:

C++

char * str;

str is a char * - a pointer to a character. But you don't assign anything to the variable before you try to copy into the memory:

C++

strcpy(str,"Hello");

Which means that the copy operation goes to a nonexistant area of memory, and you get an error.
Assign an area of memory (with malloc perhaps) and it should start to work.

[edit]I HATE MARKDOWN![/edit]

Why this is segmentation fault ?

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