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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 ...
Answer
#9: Post edited
by
Karl Knechtel
·
2023-11-21T14:31:42Z (about 1 year ago)
Use code formatting for code and list formatting for the list; fix some minor grammar etc. issues
Reasons to use the construct> #define FOO(x) do{...} while(0);(1) as mentioned above, solves the problem of> if (...) FOO(y);(2) you can to 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 which FOO(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)), where func has side effects. This includes FOO(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 calling func2(x) etc(3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct `#define FOO(x) do{...} while(0);`:
- 1. As mentioned above, doing so solves the problem of
- ```
- if (...) FOO(y);
- ```
- 1. 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 which `FOO(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))`, where `func` has side effects. This includes `FOO(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 calling `func2(x)` etc.
- 1. Without the `do{}` block, there might be ambiguity as to whether `FOO(x)` expands into a statement or an expression. `do{}` makes it clearly a statement, which can help catch someone trying to use `FOO(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.
#8: Post edited
by
Pete W
·
2020-12-17T15:42:17Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you can to 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 which FOO(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)), where func has side effects. This includes FOO(y++) etc. After the macro expands, the side effect would happening multiple times.- 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you can to 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 which FOO(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)), where func has side effects. This includes FOO(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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#7: Post edited
by
Pete W
·
2020-12-17T15:40:51Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
(2) you get to declare more 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 in the same block in which FOO(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)), where func has side effects. This includes FOO(y++) etc. After the macro expands, the side effect would happening multiple times.
- 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you can to 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 which FOO(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)), where func has side effects. This includes FOO(y++) etc. After the macro expands, the side effect would happening multiple times.
- 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#6: Post edited
by
Pete W
·
2020-12-17T15:39:53Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func(y)) , where func has side effects.The do {} block is then used 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you get to declare more 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 in the same block in which FOO(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)), where func has side effects. This includes FOO(y++) etc. After the macro expands, the side effect would happening multiple times.
- 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#5: Post edited
by
Pete W
·
2020-12-17T15:36:54Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
> if (...) FOO(x);- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
- Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func(y)) , where func has side effects.
- The do {} block is then used 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(y);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
- Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func(y)) , where func has side effects.
- The do {} block is then used 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#4: Post edited
by
Pete W
·
2020-12-17T15:36:25Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func1(y)) , where func1() has side effects. The do {} block is then used as follows:> #define FOO(x) do { long xx = x; func2(xx); func3(xx); }- Obviously there is still an assumption about type of x, but that is the case anyway if you are going to be calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
- Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func(y)) , where func has side effects.
- The do {} block is then used 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 calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#3: Post edited
by
Pete W
·
2020-12-17T15:34:30Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
(2) you get to declare more 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 in the same block in which FOO(x) gets expanded. Why declare variables inside a macro? To avoid unanticipated behavior for the caller due to macro argument getting expanded more than once. For example:> #define FOO(x) do { long xx = x; func1(xx); func2(xx); }Obviously there is still a type conversion issue, but that is the case just the same if you are going to be calling func1(x) etc(3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(x) where an expression is expected. See also: "statement-expressions", aka ({...}), which is mostly the same except the result is an expression rather than a statement.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded.
- Why declare variables inside a macro? In case someone writes FOO(y++). Or more generally FOO(func1(y)) , where func1() has side effects. The do {} block is then used as follows:
- > #define FOO(x) do { long xx = x; func2(xx); func3(xx); }
- Obviously there is still an assumption about type of x, but that is the case anyway if you are going to be calling func2(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(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.
#2: Post edited
by
Pete W
·
2020-12-17T15:29:19Z (about 4 years ago)
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
(2) you get to declare more 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 in the same block in which FOO(x) gets expanded- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(x) where an expression is expected. See also: "statement-expressions", aka ({...}), which is mostly the same except the result is an expression rather than a statement.
- Reasons to use the construct
- > #define FOO(x) do{...} while(0);
- (1) as mentioned above, solves the problem of
- > if (...) FOO(x);
- (2) you get to declare more 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 in the same block in which FOO(x) gets expanded. Why declare variables inside a macro? To avoid unanticipated behavior for the caller due to macro argument getting expanded more than once. For example:
- > #define FOO(x) do { long xx = x; func1(xx); func2(xx); }
- Obviously there is still a type conversion issue, but that is the case just the same if you are going to be calling func1(x) etc
- (3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(x) where an expression is expected. See also: "statement-expressions", aka ({...}), which is mostly the same except the result is an expression rather than a statement.
#1: Initial revision
by
Pete W
·
2020-12-17T00:01:53Z (about 4 years ago)
Reasons to use the construct
> #define FOO(x) do{...} while(0);
(1) as mentioned above, solves the problem of
> if (...) FOO(x);
(2) you get to declare more 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 in the same block in which FOO(x) gets expanded
(3) Without the do{} block, there might be ambiguity as to whether FOO(x) expands into a statement or an expression. do{} makes it clearly a statement, which can help catch someone trying to use FOO(x) where an expression is expected. See also: "statement-expressions", aka ({...}), which is mostly the same except the result is an expression rather than a statement.