Segmentation fault (core dumped) with memset in char pointer in C++

Question

I have two version of code. One works and the other doesnot.

Working code is a follows :

 int main()
    {
        int i;
        char str[] = "Hello World";
        std::cout<<"The string value before memset is : "<<str<<std::endl;
        memset (str,'-',6);
        std::cout<<"The string value after memset is : "<<str<<std::endl;
        return 0;
    }

It gives expected output :

The string value before memset is : Hello World
The string value after memset is : ------World

Now, I have another version of code in which I want to use the char pointer but this code is not working. I get the following output :

int main()
{
    char *str;
    str = "Hello World";
    std::cout<<"The string value before memset is : "<<str<<std::endl;
    memset (str,'-',6);
    std::cout<<"The string value after memset is : "<<str<<std::endl;
    return 0;
}
The string value before memset is : Hello World
Segmentation fault (core dumped)

I just could not figure on what is happening. Could you help me on this?

lol - you show the code that works, but not the code that fails! — pm100, Nov 03 '17 at 21:43
i know what it says tho - it says `char *str = "Hello World"` — pm100, Nov 03 '17 at 21:44
String literals are *constant*. You can not and should not modify them. That's why you should always use `const char*` when declaring pointers to them. — Some programmer dude, Nov 03 '17 at 21:48
@Someprogrammerdude If I use const char* then it gives me error : error: invalid conversion from ‘const void*’ to ‘void*’ [-fpermissive] — nzy, Nov 03 '17 at 22:03
@enjal That's because you're not supposed to modify the data! A literal string pointed to by a pointer, is *constant*. — Some programmer dude, Nov 03 '17 at 22:27

pm100 · Answer 1 · 2017-11-03T21:54:36.893

0

I assume you have

char *str = "Hello world";

in your failing code. The memset fails because str is a pointer to a constant string. You are not allowed to change constant strings.

In the one that works

char str[] = "Hello World";

copies the constant string into a local variable. You are allowed to change that.

edited Nov 03 '17 at 21:54

answered Nov 03 '17 at 21:49

pm100

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I have to work in the pointer version. – nzy Nov 03 '17 at 22:05
1

@enjal : Then "Yousa screwed," quoth the Gungan. You can't write into constant memory without invoking the bad-weirds of undefined behaviour. – user4581301 Nov 03 '17 at 22:28
what you mean 'i have to work with the pointer one'? – pm100 Nov 04 '17 at 00:07

score 0 · Accepted Answer · answered Nov 03 '17 at 23:08

What compiler are you using? That should not compile, because "Hello World" is a const char[12], which decays to const char*. In C++ it's not valid to assign a const char* to a char*, because you cannot lose const without a cast. Modifying const data is undefined behavior, so a crash is reasonable. (Usually string literals are placed in a read-only memory segment so it will likely crash if you write to them (but that's a compiler detail, not language.))

If you want to modify the data, you must copy the string literal into your own memory. Your first is a valid way to do that.

Segmentation fault (core dumped) with memset in char pointer in C++

2 Answers2