sizeof char* array in C/C++

Question

There are plenty of similar inquires, but in my case I don\'t understand what isn\'t working:

int mysize = 0;
mysize = sizeof(samplestring) / sizeof(*samplestrin         


        

Answer 1

It looks as though you have a pointer, not an array. Arrays are converted to pointers when the program requires it, so you'd get:

size_t size(char * p) { // p is a pointer
    return sizeof(p) / sizeof(*p); // size of pointer
}

size_t size(char p[]) { // p is also a pointer        
    return sizeof(p) / sizeof(*p); // size of pointer
}

although, since sizeof (char) == 1, the division is redundant here; if the pointer were to a larger type, then you'd get a differently unexpected result.

In C++ (but obviously not C), you can deduce the size of an array as a template parameter:

template 
size_t size(T (&)[N]) {
    return N;  // size of array
}

or you can use classes such as std::vector and std::string to keep track of the size.

In C, you can use strlen to find the length of a zero-terminated string, but in most cases you'll need to keep track of array sizes yourself.