Consider the following snippet:
请考虑以下代码段：
void f(void);
void g(…)
{
…
return f();
…
}
Is this return f(); valid according to C11?
这是返回f（）;根据C11有效吗？
I am not advocating using this pattern: if it works at all, it is obviously equivalent to f(); return; (where the return; itself would be redundant if this is at the end of function g()). I am asking this question in the context of the static analysis of C programs, where the C code has already been written by someone else and the question is deciding whether or not it is valid according to the standard.
我并不主张使用这种模式：如果它起作用，它显然等同于f（）;返回; （返回;如果它在函数g（）的末尾，则本身将是多余的）。我在C程序的静态分析的上下文中问这个问题，其中C代码已经由其他人编写，问题是根据标准判断它是否有效。
I would interpret C11 6.8.6.4:1 as meaning that it is non-standard and should be statically rejected. Is it possible to interpret it differently (I have found this pattern in actual and otherwise high-quality source code)?
我认为C11 6.8.6.4:1意味着它是非标准的并且应该被静态拒绝。是否有可能以不同的方式解释它（我在实际和其他高质量的源代码中找到了这种模式）？
Constraints
约束
A return statement with an expression shall not appear in a function whose return type is void. A return statement without an expression shall only appear in a function whose return type is void.
带有表达式的return语句不应出现在返回类型为void的函数中。不带表达式的return语句只能出现在返回类型为void的函数中。
3 个解决方案
#1
18
Anything after return is an expression.
返回后的任何东西都是表达式。
6.8.6:1 Jump statements
Syntax ... return expressionopt;
And standard says that:
标准说：
A return statement with an expression shall not appear in a function whose return type is void. ....
带有表达式的return语句不应出现在返回类型为void的函数中。 ....
f() is also an expression here. The compiler should raise a warning
f（）也是这里的一个表达式。编译器应该发出警告
[Warning] ISO C forbids 'return' with expression, in function returning void [-pedantic]
#2
10
This clearly is a constraint violation, in particular in view of
这显然是违反约束的行为，特别是考虑到这一点
6.3.2.2 void: The (nonexistent) value of a void expression (an expression that has type void) shall not be used in any way,
6.3.2.2 void：void表达式（具有void类型的表达式）的（不存在）值不得以任何方式使用，
That means that the incomplete type void is a dead end that cannot be reused for any purpose whatsoever.
这意味着不完整类型的空洞是一个死胡同，不能为任何目的重复使用。
#3
0
It clearly states A return statement without an expression shall only appear in a function whose return type is void, try and execute this:
它清楚地说明没有表达式的return语句只出现在返回类型为void的函数中，请尝试执行：
void g()
{
return; // does not return any expression at all
}
void f()
{
return g();
}
int main(void) {
f();
return 0;
}
#1
18
Anything after return is an expression.
返回后的任何东西都是表达式。
6.8.6:1 Jump statements
Syntax ... return expressionopt;
And standard says that:
标准说：
A return statement with an expression shall not appear in a function whose return type is void. ....
带有表达式的return语句不应出现在返回类型为void的函数中。 ....
f() is also an expression here. The compiler should raise a warning
f（）也是这里的一个表达式。编译器应该发出警告
[Warning] ISO C forbids 'return' with expression, in function returning void [-pedantic]
#2
10
This clearly is a constraint violation, in particular in view of
这显然是违反约束的行为，特别是考虑到这一点
6.3.2.2 void: The (nonexistent) value of a void expression (an expression that has type void) shall not be used in any way,
6.3.2.2 void：void表达式（具有void类型的表达式）的（不存在）值不得以任何方式使用，
That means that the incomplete type void is a dead end that cannot be reused for any purpose whatsoever.
这意味着不完整类型的空洞是一个死胡同，不能为任何目的重复使用。
#3
0
It clearly states A return statement without an expression shall only appear in a function whose return type is void, try and execute this:
它清楚地说明没有表达式的return语句只出现在返回类型为void的函数中，请尝试执行：
void g()
{
return; // does not return any expression at all
}
void f()
{
return g();
}
int main(void) {
f();
return 0;
}