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Comments on What is do { } while(0) in macros and should we use it?
Parent
What is do { } while(0) in macros and should we use it?
Background
I can see the need to use { } when implementing a function-like macro such as this one:
#define HCF(code) fprintf(stderr, "halt and catch fire"); exit(code);
Because if we use the following calling code
bool bad = false;
if(bad)
HCF(0xDEADBEEF);
printf("good");
Then it will expand to
bool bad = false;
if(bad)
fprintf(stderr, "halt and catch fire");
exit(0xDEADBEEF);
printf("good");
which was not the intention - it will now exit with an error code even though the program is working as intended.
So the correct way would be to write the macro like this instead:
#define HCF(code) { fprintf(stderr, "halt and catch fire"); exit(code); }
Question
But I've come across macros that instead of just plain braces { } uses this:
#define HCF(code) do { fprintf(stderr, "halt and catch fire"); \
exit(code); } while(0)
This seems similar to { } but more obscure. What's the purpose of this do { } while(0) and why should it be used instead of { }?
Post
Reasons to use the construct #define FOO(x) do{...} while(0);:
-
As mentioned above, doing so solves the problem of
if (...) FOO(y); -
You can declare variables inside the block, and they harmlessly go out of scope at the end of the block. Without the
do{}block, the same symbol would cause a warning (or worse) if it were declared elsewhere in the block in whichFOO(x)gets expanded.Why declare variables inside a macro? In case a macro argument appears more than once in the body, and someone writes
FOO(func(y)), wherefunchas side effects. This includesFOO(y++)etc. After the macro expands, the side effect would happen as many times as the macro argument appears in the body!Declaring a variable inside a do {} block prevents this, as follows:
#define FOO(x) do { long xx = x; func2(xx); func3(xx); } while (0);Obviously there is still an assumption about type of
x, but that is the case anyway if you are going to be callingfunc2(x)etc. -
Without the
do{}block, there might be ambiguity as to whetherFOO(x)expands into a statement or an expression.do{}makes it clearly a statement, which can help catch someone trying to useFOO(x)where an expression is expected. See also: "statement-expressions", aka({...}), which is a construct in case you want the block to be an expression rather than a statement.
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