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无法在Linux中使用curl编译 - 未定义的引用

[英]Can't compile in Linux with curl - undefined reference

I'm trying to compile a c++ sample in Linux(ubuntu) using curl lib but I'm getting undefined reference to 'curl_easy_init' 我正在尝试使用curl lib在Linux(ubuntu)中编译c ++示例但是我得到了对'curl_easy_init'的未定义引用

Compile command: 编译命令:

gcc -L/usr/local/lib -lcurl -I/usr/local/include -o request request.cpp

result: 结果:

/tmp/ccZwDiCf.o: In function 'main':<br>
request.cpp:(.text+0xa): undefined reference to 'curl_easy_init'<br>
request.cpp:(.text+0x31): undefined reference to 'curl_easy_setopt'<br>
request.cpp:(.text+0x3d): undefined reference to 'curl_easy_perform'<br>
request.cpp:(.text+0x54): undefined reference to 'curl_easy_strerror'<br>
request.cpp:(.text+0x7b): undefined reference to 'curl_easy_cleanup'<br>
collect2: ld returned 1 exit status

Code: 码:

#include <curl/curl.h>
#include <stdio.h>

int main(int argc, char* argv[]){
    CURL *curl;
    CURLcode res;

    curl = curl_easy_init();
    if(curl) {
        curl_easy_setopt(curl, CURLOPT_URL, "http://google.com");
        res = curl_easy_perform(curl);
        if(res!=CURLE_OK) 
            fprintf(stderr, "curl_easy_perform() failed: %s\n",curl_easy_strerror(res));
        curl_easy_cleanup(curl);
    }
    return 0;
}

Answer just to close this ask. 回答只是为了结束这个问题。

For resumate, you had to place the -lcurl option at the end of your compilation command line. 对于resumate,您必须将-lcurl选项放在编译命令行的末尾。

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