I've asked the same question about strncpy
, but there the string
ends up containing the whole input string. When passing a string to vsnprintf
the last character always gets chopped off: https://rextester.com/UIQMX91570
For simplicity I've also included the live example link above inline in the code:
void bar(const char* format, va_list vlist) {
const auto buf_size = vsnprintf(nullptr, 0U, format, vlist);
string buffer(buf_size, '\0');
vsnprintf(data(buffer), buf_size, format, vlist);
cout << data(buffer) << endl;
}
void foo(const char* format, ...) {
va_list vlist;
va_start(vlist, format);
bar(format, vlist);
va_end(vlist);
}
If I call this with: foo("lorem ipsum %d", 13)
the output I get is:
lorem ipsum 1
Where as I would have expected: lorem ipsum 13
Can anyone explain the discrepancy? When I debug I get a buf_size
of 14 which should be enough to contain the entire string, yet it does not :(
Can anyone explain the discrepancy?
Because their documented behavior is different.
strncpy()
If count is reached before the entire array src was copied, the resulting character array is not null-terminated.
but vsnprintf()
At most buf_size-1 characters are written. The resulting character string will be terminated with a null character, unless buf_size is zero.
emphasis is mine.
Because man page clearly says that
If the output was truncated due to this limit then the return value is the number of characters (not including the trailing '\0') which would have been written to the final string if enough space had been available.
If you'd check the return value of your second vsnprintf
call, you'd see that return value is equal to the size, as in the man page:
Thus, a return value of size or more means that the output was
truncated.
The buf_size
parameter to vsnprintf
specifies how many characters to write, including the terminating NUL character. The return value is the number of characters produced, not including the terminating NUL character.
You want
const auto buf_size = vsnprintf(nullptr, 0U, format, vlist) + 1;