-W 标志，用于 gcc 检测获取未初始化变量的 const 指针

Question

我有以下代码，保存在 Coding Ground here:

#include <stdio.h>

void foo(int const *x)
{
    printf("Hello, %d!\n", *x);
}

int main()
{
    int y;

    foo(&y);
    y = 3;
    printf("Hello, World %d!\n", y);
    return 0;
}

我编译：

gcc -std=c99 -Wall -Wextra -Wuninitialized -o main *.c

但是，我没有收到关于获取未初始化变量的 const 指针的警告，并且我 cannot find a suitable flag.

注意：将代码粘贴到 Gimpel's Online Lint 中得到预期的结果：

test.c 12 Warning 603: Symbol 'y' (line 10) not initialized

Answer 1

-Wuninitialized Warn whenever an automatic variable is used without first being initialized.

These warnings are possible only in optimizing compilation, because they require data-flow information that is computed only when optimizing. If you don't specify `-O', you simply won't get these warnings.

如果您尝试编译：

gcc -std=c99 -Wall -Wextra -Wuninitialized -O2 -o main *.c

警告将是：

pippo.c: In function ‘main’:
pippo.c:56:5: warning: ‘y’ is used uninitialized in this function [-Wuninitialized]
     printf("Hello, %d!\n", *x);
     ^
pippo.c:61:9: note: ‘y’ was declared here
     int y;
         ^

Answer 2

为什么这里应该有问题？您只是通过该函数传递了该变量，该函数将指针作为参数。你没有传递它的值，你用 & 传递了它的地址所以编译器不会认为这是错误的，即使你稍后尝试打印它的垃圾值。请记住，您此时打印的是指针 (*x) 的值，而不是变量本身。如果你不自己做，变量不是用垃圾值初始化的吗？

但是如果你不使用指针:

#include <stdio.h>

void foo(int const x)
{
    printf("Hello, %d!\n", x);
}

int main(void){
    int y;

    foo(y);
    y = 3;
    printf("Hello, World %d!\n", y);
    return 0;
}

现在情况有所不同，编译器知道这一点：

 program.c:12:5: error: ‘y’ is used uninitialized in this function [-Werror=uninitialized]
     foo(y);
     ^
cc1: all warnings being treated as errors

-W 标志，用于 gcc 检测获取未初始化变量的 const 指针

-W flag for gcc to detect taking const pointer of uninitialised variable

c

gcc-warning