c99

来源： https://stackoverflow.com/questions/46233400/does-union-support-flexible-array-members

来源： https://stackoverflow.com/questions/46233400/does-union-support-flexible-array-members

来源： https://stackoverflow.com/questions/46233400/does-union-support-flexible-array-members

来源： https://stackoverflow.com/questions/44590043/why-is-generic-keyword-supported-in-c99-or-c90-modes

来源： https://stackoverflow.com/questions/53380203/why-are-uintptr-t-and-intptr-t-optional-types-in-the-c-and-c-standard

问题 C99 introduced a new function argument notation where the static keyword can be used to specify that the argument has at least N elements. 6.7.6.3 Function declarators, p7 A declaration of a parameter as ''array of type'' shall be adjusted to ''qualified pointer to type'', where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function,

问题 C99 introduced a new function argument notation where the static keyword can be used to specify that the argument has at least N elements. 6.7.6.3 Function declarators, p7 A declaration of a parameter as ''array of type'' shall be adjusted to ''qualified pointer to type'', where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function,

问题 C99 introduced a new function argument notation where the static keyword can be used to specify that the argument has at least N elements. 6.7.6.3 Function declarators, p7 A declaration of a parameter as ''array of type'' shall be adjusted to ''qualified pointer to type'', where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function,

问题 1) Why is the macro MSG not expanded in the following expression? #define MSG Hello #define HELLO(name) MSG ## name void HELLO(Dave) () {} Using gcc -E -P test.cpp Output: void MSGDave () {} MSG name expands to Hello Dave . And MSG # name expands to Hello "Dave" . So what causes gcc not to expand MSG ## name ? 2) Is there a workaround? Is there a preprocessor directive like defined(x), such as expand(x)? 回答1: #define MSG Hello #define cat(x, y) x ## y #define cat2(x, y) cat(x, y) #define

问题 I just read this: In C++ (and C99), we can pass by reference, which offers the same performance as a pointer-pass. So I tried this simple code: #include <stdio.h> void blabla(int& x){ x = 5; } int main(){ int y = 3; printf("y = %d\n", y); blabla(y); printf("y = %d\n", y); } The output was: gcc test.c -o test -std=c99 test.c:3:16: error: expected ';', ',' or ')' before '&' token test.c: In function 'main': test.c:10:2: warning: implicit declaration of function 'blabla' Now I'm confused. Is