C中的分段错误将字符串分配给char

Question

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char a="9jhjhi";
    printf("%s",a);
}

Why does this throw a segmentation fault? What happens behind the screens?

Answer 1

You need to use char *a = "..." .

printf when passed %s will run through a string looking for a 0 / NULL byte. In your case you are not assigning a string literal to a , and in fact your compiler should have thrown a warning that you were trying to initialize a char from a pointer.

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char *a="9jhjhi";
    printf("%s",a);
}

Answer 2

Your error is that

char a="9jhjhi";

should be

char *a="9jhjhi";

What happens is undefined behavior - so anything could happen.

Your assigning a string literal to a char, so your a will contain a pointer(to the beginning of that string) converted to a char - whatever that'll be.

%s conversion in printf assumes you pass it a string, which must be a char* pointing to a sequence of chars ending with a 0 terminator. You passed it a char, which certainly does not meet those requirements, so it's quite undefined what'll happen - a crash could be common.

You should also return something from the main() method - it's declared to return an int after all.

Answer 3

C does not have strings as in String b = new String();

C has arrays of type char.

So char a="123123" should be a character array.

You aren't using anything from stdlib.h in that code either so there is no reason to #include it.

Edit: yeah, what nos said too. An array name is a pointer.

Answer 4

a is initialized to a (cast to integer and truncated because char is 3 or 7 bytes too small) pointer that points to a char array (propably somewhere in ROM). What follows is undefined, but it's propably like this: When you pass it to printf with a %s in the format string, it takes the value of a (something in 0-255 ) and 3 (or 7) unrelated bytes from the stack, gets some bogus address and wreaks havok by accessing someone else's memory.

Use char *a = ... .

Answer 5

You mean

char *a = "9jhjhi";

If this compiles without warnings, your compiler settings are messed up. The warnings from gcc show plainly what's happening:

test.c: In function ‘main’:
test.c:5: warning: initialization makes integer from pointer without a cast
test.c:6: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘int’

The string literal is interpreted as a pointer, which is converted (and truncated) to a char .

Then the char is sent into printf as a number, but interpreted as a string. Since it's not null-terminated in any way, printf probably overruns memory when racing through that "string".

Answer 6

When you declare char a without a pointer symbol, you are only allocating enough space on the stack for a single character.

Strings in C are represented by a char array, or a pointer to a char array, terminated by the null character '\\0' . But if you use a literal string, the compiler takes care of all of that for you.

Here's how you can get your code to work, then:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char *a = "9jhjhi";
    printf("%s",a);
}

Answer 7

First of all, you're trying to save an entire string into a single char variable. Use char array (char[size]) instead. You may also want to terminate the string with "\\0".

Answer 8

You could remove this error in two ways.

1.char * p="karthik A"

2.char [ ]p="karthik A"