对 strncpy_s() 使用 destsz = sizeof(dst) 是否有缺点？

Question

I'm working off this: https://en.cppreference.com/w/c/string/byte/strncpy

The function header is described as this:

errno_t strncpy_s(char *restrict dest, rsize_t destsz, const char *restrict src, rsize_t count);

Is there any significant drawback other than speed, to setting destsz = sizeof(dest) ?

Answer 1

If you were to use strncpy_s like this

strncpy_s(dest, sizeof(dest), src, sizeof(src));

This assumes that dest and src are expressions representing an array variable, like this.

char a[3];
char b[4] = {'a', 'b', 'c', '\0'};
strncpy_s(a, sizeof(a), b, sizeof(b));

If instead, dest or src were pointers (for example, char* ), like this.

char* ptr1 = a;
char* ptr2 = b;
strncpy_s(ptr1, sizeof(ptr1), ptr2, sizeof(ptr2));

The code would compile, but the values would be wrong, because sizeof(ptr1) would be the size of the pointer object, not the size of the array. In that case, you would have to provide the size in some other way

So, no direct drawbacks, but using this pattern can be error prone.

As suggested by @Alex, when using the Microsoft C Runtime Library, you can directly pass the array expression for the destination without passing the size separately

strncpy_s(a, ptr2, sizeof(b));

Then, providing the size is not necessary, and doing so brings nothing except potentially mixing up an array and a pointer. For example, if one accidentally writes strncpy_s(ptr1, b, sizeof(b)) , the code would not compile, rather than work with the wrong values

Answer 2

No, in fact the template overloads for strncpy_s do exactly that when you provide a fixed width array as the destination parameter. However, sizeof(dest) will not work if dest is a pointer type, as the size of a pointer is always the size of the native machine word (32 or 64 bit). If you have a pointer, you will need to keep track of the allocated size in a second variable.