C ++不能返回变量

时间:2011-04-27 18:04:14

标签: c++ variables return main

嘿所有人!这段代码总是返回0,即使errcheck将具有非零值。如果我使用return 1;它按预期工作。请帮帮忙?

   int errcheck = system(docommand.c_str());
    if (errcheck != 0)
    {
        cerr << "Could not retrieve tarball!" << " Errcheck status (debug): " << errcheck << endl;
        return errcheck;
    }

以下是完整代码:

#include <iostream>
#include <sys/stat.h>
#include <sys/types.h>
#include <stdlib.h>
#include <string>
using namespace std;

int main(int argc, char* argv[])
{
    umask(0);
    mkdir("/tmp/.aget", 0755);
    chdir("/tmp/.aget");

    for (int i = 1; i < argc; i++)
    {

        string target(argv[i]);
        string docommand("");
        string s1("wget -q http://aur.archlinux.org/packages/");
        string s2("/");
        string s3(".tar.gz");
        docommand += s1;
        docommand += target;
        docommand += s2;
        docommand += target;
        docommand += s3;
        cout << "Downloading AUR tarball for '" << target << "'..." << endl;
        int errcheck = system(docommand.c_str());
        if (errcheck != 0)
        {
            cerr << "Could not retrieve tarball!" << " Errcheck status (debug): " << errcheck << endl;
            return errcheck;
        }
    }

    for (int i = 1; i < argc; i++)
    {
        string target(argv[i]);
        string docommand("");
        string s1("tar xf ");
        string s2(".tar.gz");
        docommand += s1;
        docommand += target;
        docommand += s2;
        cout << "Extracting '" << target << ".tar.gz'..." << endl;
        system(docommand.c_str());
    }

    for (int i = 1; i < argc; i++)
    {
        string target(argv[i]);
        string docommand("");
        chdir("/tmp/.aget");
        chdir(target.c_str());
        system("makepkg -csim --noconfirm > /dev/null");
    }

    rmdir("/tmp/.aget");

    return 0;
}

2 个答案:

答案 0 :(得分:1)

Unix退出状态限制为值0-255,即无符号8位整数的范围。因此,你看不到2048。

有关详细信息,请参阅Exit Status wiki页面。

答案 1 :(得分:0)

我怀疑wget总是返回0.

这是因为http请求的实际错误状态在流中。