why segmentation fault !

wbgxx

I want to swap two string

#include <stdio.h>
#include <string.h>
void swap(char *a1, char *a2)
{
char ch[20];
if (a1 != NULL || a2 != NULL)
{
strcpy(ch, a1);
strcpy(a1, a2);
strcpy(a2, ch);

	}
else
	return ;

}

int main()
{
char *a1 = "hello";
char *a2 = "world";
printf("Before the swap a1=%s, a2=%s", a1, a2);
swap(a1, a2);
printf("After the swap a1=%s, a2=%s", a1, a2);
return 0;
}

Aescleal

Remember what I said about the difference between pointers and arrays earlier? What's probably happening is that the arrays your pointers are pointing to are in a chunk of memory in the program's text section. This section is usually read only - you don't want programs modifying their own code as a rule. The first strcpy will work okay - you're copying data from the text section to an array on the stack, which is read/write. when you do the second strcpy you're copying from the text section to another bit of the text section and BOOM says the operating system. If you declare a1 and a2 as:

char a1[] = "hello";
char a2[] = "world";

it'll start working if I'm correct. Another way of looking at it is...

char a[] = "Hello";

translates to: - allocate enough memory on the stack to hold 6 characters - copy "Hello\0" into the memory while:

char *a = "Hello";

translates to: - allocate enough memory on the stack to hold a pointer - initialise the pointer with the address of the constant string "Hello" in the text section Incidentally this being able to use a char * to point at a const block of characters initialised at compile time is a legacy of the original K&R C which didn't have things like const - it's one of the irritating bits of programming in C. Unfortunately you just have to get used to it. Hope that helps, Ash

Fareed Rizkalla

Use & with the * in swap. char*& a1

wbgxx

thanks ! but I still have question why this can be run

#include <iostream.h>
#include <cstring>
void swap(char *,char *);
int main()
{char str1[20]="hello",str2[20]="how are you ";
char pt[20];
cout<<"str1"<<" "<<str1<<endl;
cout<<"str2"<<" "<<str2<<endl;
strcpy(pt,str1);
strcpy(str1,str2);
strcpy(str2,pt);
cout<<"str1"<<" "<<str1<<endl;
cout<<"str2"<<" "<<str2<<endl;
return 0;
}

Lost User

Your problem seems to be obvious.You are not allowed to modify string contents if they are allocated like a1 and a2.You should probably change your code like that:

char a1\[20\];
strcpy(a1,"Hello");	

char a2\[20\];
strcpy(a2,"world");

In addition there was almost the same thread recently.

Life is a stage and we are all actors!

wbgxx

wonderful！ I see， thanks in advance！

Aescleal

That works because you're using arrays - you're saying "allocate 20 bytes on the stack and initialise it with this string." With the original version you were saying "Initialise this pointer to point at a chunk of read only memory." The difference is where the data your swapping is in memory. Have a look at the address of your variables in a debugger. You'll probably see all three having similar addresses in the second version. In the first version the a1 and a2 will have similar addresses and a lot different to that of the swap temporary. Cheers, Ash

Niklas L

There is another access violation waiting to happen in your swap function. None of the arguments in a call to strcpy can be null. You need to check for a1 != NULL && a2 != NULL. Also, watch out for the buffer overruns you might encounter if a1 is longer than 19 characters, or the allocated buffers of a2 and a1 are of different sizes.

home

Aescleal

You don't need the explicit strcpy call, the compiler will generate the code to initialise the array properly if you just use:

char a[ 20 ] = "Hello";

Ash

David Crow

Read here.

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Aescleal

Thanks for posting that, the piccies work far better than my text. Ash