使用-std = c99在gcc中嵌套函数

Question

From what I am reading the below code is invalid c99, however I seem to be able to compile it using gcc -std=c99 which, to my knowledge should disable the GNU extension that allows for embedded functions. I cannot seem to figure out why this is the case.

int main() {
    int e() {
        printf("testing");
        return 0;
    };

    e();
    return 0;
}

Answer 1

In order to get warnings on non-conformant code you need to use the -pedantic flag as well, then you will see the following ( see it live ):

warning: ISO C forbids nested functions [-Wpedantic]
 int e() {
 ^

To turn this into an error you can use -Werror to turn warnings into errors or -pedantic-errors .

From the gcc docs on standards support :

to obtain all the diagnostics required by the standard, you should also specify -pedantic (or -pedantic-errors if you want them to be errors rather than warnings)