为什么我收到"没有效果的声明"警告？

Question

以下是我的代码

/* Initialise default without options input. */
options -> processHiddens = false;
options -> timeResolution = DEFAULT_MOD_TIMES;
options -> performSync = true;
options -> recursive = false;
options -> print = false;
options -> updateStatus = true;
options -> verbose = false;
options -> programname = malloc(BUFSIZ);
options -> programname = argv[0];

while ((opt = getopt(argc, argv, OPTLIST)) != -1)
{
    switch (opt)
    {
        case 'a':
            !(options -> processHiddens);
        case 'm':
            options -> timeResolution = atoi(optarg);
        case 'n':
            !(options -> performSync);
        case 'p':
            !(options -> print);
        case 'r':
            !(options -> recursive);
        case 'u':
            !(options -> updateStatus);
        case 'v':
            !(options -> verbose);
        default:
            argc = -1;
    }
}

我想要做的是每次输入一个选项时都要翻转布尔语句,因此做类似的事情

!(options -> processHiddens);

而不仅仅是

options -> processHiddens = true;

但是在编译时我收到以下警告:

mysync.c: In function ‘main’:
mysync.c:32: warning: statement with no effect
mysync.c:36: warning: statement with no effect
mysync.c:38: warning: statement with no effect
mysync.c:40: warning: statement with no effect
mysync.c:42: warning: statement with no effect
mysync.c:44: warning: statement with no effect

Answer 1

因为!(options -> processHiddens)是表达式,并且您没有将结果分配给任何东西.你需要这样的东西:

options->processHiddens = !options->processHiddens;

Answer 2

因为!(options -> processHiddens);"和"一样" 40 + 2.它真的没有效果:-)

printf("foo");
40 + 2;
printf("bar");

你要

option -> processHiddens = !(options -> processHiddens);
break;        /* without break, all the following lines will execute */