Question

WARNING: While resolving call to function 'help' arguments were dropped!

如果我使用gcc -O3 codice -o out/codice进行编译，我会收到这个令人讨厌的警告。它完成了编译，所以我只是想知道为什么会这样。

我在安装xcode 4.3的狮子上安装了mac。它使用的编译器是i686-apple-darwin11-llvm-gcc-4.2

请注意，如果我将-O3切换为-O2 / -O1 / -O，则会继续发出相同的警告

代码如下....

#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <fcntl.h>
#include <signal.h>
#include <string.h>
#include <math.h>

void help();

char *codice_encrypt(char *in);
char *codice_decrypt(char *in);

void sigint_handler(int signal) {
    exit(2);
}

int main (int argc, char **argv) {

    if (argc != 3) {
        help();
    }

    signal(SIGINT, sigint_handler);
    char *ret, *status;
    int tmpret;

    if (strcmp(argv[1], "e") == 0 || strcmp(argv[1], "encrypt") == 0) {
        status = "Encrypting...\n";
        tmpret = write(STDOUT_FILENO, status, strlen((const char *)status));
        ret = codice_encrypt(argv[2]);
    } else if (strcmp(argv[1], "d") == 0 || strcmp(argv[1], "decrypt") == 0) {
        status = "Decrypting...\n";
        tmpret = write(STDOUT_FILENO, status, strlen((const char *)status));
        ret = codice_decrypt(argv[2]);
    } else {
        status = "Could not understand command line arguments O.o\n";
        tmpret = write(STDOUT_FILENO, status, strlen((const char *)status));
        help(STDOUT_FILENO);
    }

    status = "Success!\nRetval:\n\n";
    tmpret = write(STDOUT_FILENO, status, strlen((const char *)status));
    tmpret = write(STDOUT_FILENO, ret, strlen((const char *)ret));
    tmpret = write(STDOUT_FILENO, "\n\n", 2);

    return 0;

}

void help() {
    int tmpret;
    const char *logo = "\n\n\t\tooooooo__oo______________________\n"
                 "\t\t_____oo______oo_oo_oo__oo_oo_oo__\n"
                 "\t\t____oo___oo__ooo_oo__o_ooo_oo__o_\n"
                 "\t\t___o_____oo__oo__oo__o_oo__oo__o_\n"
                 "\t\t_oo______oo__oo__oo__o_oo__oo__o_\n"
                 "\t\tooooooo_oooo_oo______o_oo______o_\n"
                 "\t\t_________________________________\n\n\n";
    tmpret = write(STDOUT_FILENO, logo, strlen(logo));
    const char *help = "Help\n"
                       "Is\n"
                       "Still\n"
                       "Being\n"
                       "Written :)\n\n";
    tmpret = write(STDOUT_FILENO, help, strlen(help));
    exit(1);
}

char *codice_encrypt(char *in) {
    unsigned int i;
    char ya = 0x55;
    write(STDOUT_FILENO, "0x", 2);
    char itret[50];
    sprintf(itret,"%#hhx",ya);
    write(STDOUT_FILENO, itret, strlen(itret));
    write(STDOUT_FILENO, "\n", 1);
    ya = ya & 0x0F;
    ya = ya >> 1;
    sprintf(itret,"%#hhx",ya);
    write(STDOUT_FILENO, itret, strlen(itret));
    write(STDOUT_FILENO, "\n", 1);
    ya = 0x55;
    ya = ya & 0xF0;
    ya = ya << 1;
    sprintf(itret,"%#hhx",ya);
    write(STDOUT_FILENO, itret, strlen(itret));
    write(STDOUT_FILENO, "\n", 1);
    return in;
}

char *codice_decrypt(char *in) {
    return in;
}

谢谢！

Answer 1

这里help（）有可变数量的参数（我假设我们正在谈论C代码）。但是，没有传递任何参数，也没有读取参数。其中一个优化通过实现了这一点并发出了这样的警告。将help（）定义为help（void）将解决问题。

为什么在使用优化进行编译时会收到此警告？

1 个答案: