Use of sizeof with an expression as operand CXX-W1188
Anti-pattern
Minor
The operator sizeof never evaluates the expression provided as an argument
unless it's a variable length array. It only determines the type of the operand.
(See reference).
When an expression, like a function invocation, is passed as an argument to
sizeof, the function is never called. We can say the same about an expression
like z = x + y.
Such an expression should be evaluated and stored in a variable before passing its result to sizeof.
Bad practice
#include <stdio.h>
char get_option(void) {
char option[2];
printf("Choice: ");
if (fgets(option, 2, stdin) == NULL)
return (char)0;
return option[0];
}
int main() {
// In the following line the function `get_option` will never be called and
// operator `sizeof` will only evaluate the type of the expression i.e. type char
size_t s = sizeof(get_option());
// Use s and opt as needed
return 0;
}
Recommended
#include <stdio.h>
char get_option(void) {
char option[2];
printf("Choice: ");
if (fgets(option, 2, stdin) == NULL)
return (char)0;
return option[0];
}
int main() {
// Call the function `get_option` and invoke operator `sizeof` the return
// value
char opt = get_option();
size_t s = sizeof(opt);
// Use s and opt as needed
return 0;
}
Reference
6.5.3.4 [The sizeof and _Alignof operators] -- See point 2 under Section 6.5.3.4