Why isn't working null pointer check for C style string?

Question

This is my function:


void printCString(char *s)
{
    while (s != nullptr) // printing doesn't stop after ! from passed string.
    {
        std::cout << *s;
        ++s;
    }
}

and I call it:

char s[]{ "Hello, world!" };
printCString(s);

If I replace stop condition from while block with:

while (*s != '\0')

than it's working well. Can anybody explain me why this behavior?

Where are you stuck exactly? While traversing a C-style string you'll eventually find a zero character, but the pointer you use for traversal has no reason to become null. — Quentin
– Quentin, Commented Dec 16, 2019 at 9:46
s points somewhere in memory. You are checking whether s points to memory location zero (more or less). There's no reason why the next memory location after the ! of the string should be memory location zero. — Max Langhof
– Max Langhof, Commented Dec 16, 2019 at 9:48
You basically expect s to point to nullptr when you reach the 15th iteration, but you don't know what's in there (undefined) because this memory is not part of s. — SubMachine
– SubMachine, Commented Dec 16, 2019 at 9:50

Bathsheba · Accepted Answer · 2019-12-16 09:49:02Z

4

s is never nullptr, since nullptr is unattainable via pointer arithmetic.

Conceptually you'd need to deference s, but *s != nullptr will not compile. That's no bad thing since there is no guarantee that nullptr is the same as the C-style string terminator NUL.

edited Dec 16, 2019 at 9:49

answered Dec 16, 2019 at 9:46

Bathsheba

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12 Comments

n. m. could be an AI Over a year ago

*s != nullptr shouldn't compile.

Bathsheba Over a year ago

@n.'pronouns'm.: Indeed (although I did once use a compiler that allowed it ;-) )

Bathsheba Over a year ago

@olivecoder: No, that's undefined behaviour. Pointer arithmetic is only valid within arrays.

Bathsheba Over a year ago

@olivecoder: Yup, I use one that allows it too. But on other occasions a compiler might optimise away the loop, or eat your cat.

Max Langhof Over a year ago

@olivecoder No, nullptr is not attainable via pointer arithmetic. That gcc compiles your code (which has undefined behavior) to return 0; is absolutely not proof of your theory - you can't prove that a program is valid/UB-free by showing how a compiler accepts it. And for the record, clang will turn it into an endless loop (which, btw, is also UB): godbolt.org/z/Rm5HEh How do you explain that?

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