c - 死 C 变量

Question

我正在处理一些我无法展示的私有(private)代码，但我已经制作了一些示例代码来描述我的问题:

主.c:

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <stdint.h>

typedef uint32_t (*FuncPtr)(char *s, uint32_t n);

uint32_t doSomething(char *s, uint32_t n){ 
    printf("n is %d in doSomething\n", n);
    printf("s is '%s' in doSomething\n\n", s);
    return 0;

}

uint32_t func1(char *s, uint32_t n){
    printf("n is %d in func1\n", n);
    printf("s is '%s' in func1\n\n", s);
    return doSomething(s, n);
}

uint32_t func2(char *s, uint32_t n){
    printf("n is %d in func2\n", n);
    printf("s is '%s' in func2\n\n", s);
    return doSomething(s, n);
}

uint32_t func3(char *s, uint32_t n){
    printf("n is %d in func3\n", n);
    printf("s is '%s' in func3\n\n", s);
    return doSomething(s, n);
}

void perform(FuncPtr fp, char *s, uint32_t n){
    printf("fp is location in %p\n", fp);
    printf("n is %d in perform\n", n);
    printf("s is '%s' in perform\n\n", s);
    fp(s, n);
}

int main(void) {

    srand(time(NULL));
    uint32_t r = (uint32_t)(rand() % 3 + 1);
    uint32_t n = (uint32_t)(rand() %100);
    char *s = "some string here";

    printf("r is %d\n", r);
    printf("n is %d in main\n", n);
    printf("s is '%s' in main\n\n", s);

    switch(r)
    {
        case 1:
            perform(func1, s, n);
            break;
        case 2:
            perform(func2, s, n);
            break;
        case 3:
            perform(func3, s, n);
            break;
    }

    return 0;
}

这是调用堆栈:

 0  func1(s=0x9abcdef0, n=0 << Dead >>) [..\main.c:9,1]
 1  perform(fp=0x12345678, s=0x9abcdef0, n=42)  [..\main.c:21,1]
 2  main()  [..\main.c:25,10]

正如您在顶部看到的，传递到 n 的 func1 是 << dead >> 并且与 perform 中的值不匹配，即使 s 指针始终保持在同一地址。为什么会这样？我还没有找到关于什么是死变量的有用描述。我是否可以对此代码进行任何修复，或者这是否涉及我无法在此处实现的更多编译器/内存问题？

Answer 1

死变量是被写入但永远不会被再次读取的变量。编译器可能会通过删除此代码进行优化，因为无论如何它都不会影响结果。