gdoug gdoug - 1 year ago 96
C Question

Does C's ternary conditional operator create a new scope?

The title basically says it all, but I am specifically interested to know if the following (slightly abusive) macro should be expected to work (with C99 or greater), that is, that it will return a pointer to a valid section of stack allocated memory based on the size of the expression:

#include <stdint.h>

#define NASTY(expr) ( \
8 == sizeof(expr) ? (void *)(&(uint64_t){(expr)}) : \
( \
4 == sizeof(expr) ? (void *)(&(uint32_t){(expr)}) : \
( \
2 == sizeof(expr) ? (void *)(&(uint16_t){(expr)}) : (void *)(&(uint8_t){(expr)}) \
) \
) \


The reason I am interested in such a macro is (1) I am a a nerd and (2) I am writing some code-generating macros for an embedded application that need to dynamically
the result of an expression (such as
). We are working in an energy sensitive application where function calls matter.

R.. R..
Answer Source

I don't know what you mean by "a new scope" - in C, "scope" is relevant only to identifiers, not object lifetime. Per Compound literals,

If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.

Thus, if you use this macro inside a function body, the lifetime of the pointed-to object will persist until execution of the enclosing block ends. This might or might not meet your needs; you could get in trouble by writing:

if (foo) {
    p = NASTY(bar);
/* ... */
/* Do something with p */
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