CODEWITHSUNDEEP
c
mezda
mezda

Reputation: 3647

getting segmentation fault in a small c program

i wrote a small prog :

  1 #include<stdio.h>  
  2 main(){  
  3         char* str = "string";  
  4         *str = 'k';  
  5         printf("string is = %s\n",str);
  6 }

This program gets compiled without any error or warnings, but when i run it, it gives segmentation fault. While if i rewrite the 3rd line as : char str[] = "string"; this program works perfectly fine. Can anyone please let me know what is the issue here ??? What is the difference between these two different ways of writing the 3rd line. Any help will be greatly appreciated. thanks.

Upvotes: 1

Views: 214

Answers (3)

Aryan
Aryan

Reputation: 143

I also executed the given code..

  1 #include<stdio.h>  
  2 main(){  
  3         char* str = "string";  
  4         *str = 'k';  
  5         printf("string is = %s\n",str);
  6 }

 But there is no such error. 

 Following output is shown....

 String is=Ktring

 A segmentation fault, bus error, or access violation is generally an attempt 
 to access memory that the CPU cannot physically address. So in my opinion it's   
 completely depends on execution.

 thanks...

Upvotes: 0

Duck
Duck

Reputation: 27562

char* str = "string";

This puts the string in read-only memory. It is undefined behavior (usually unpleasant behavior) when you try to modify it with the next line. Try something like

char str[] = "string";

instead.

Upvotes: 7

Matteo Italia
Matteo Italia

Reputation: 126867

char * str = "string";

declares a pointer which points to an area of memory where the string "string" is stored; it's undefined behavior to write in this area of memory, and it usually results in a crash.

Instead, you should use:

char str[]="string";

which declares a string local to your function, that is initialized to the value "string". Since the memory you are using for this string is local to your procedure you can alter it however you want.

Upvotes: 2

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