使用C ++编译和链接curl easy界面

Question

上下文

尝试运行此curl_easy_init -在我的C ++程序中启动libcurl easy会话 。 curl库有很多选择的API，包括简单的帖子和许多其他的，但我仍然坚持正确配置我的编译和链接步骤。 我查看了curl网站教程，获取有关编译和链接此库到我的程序的指导，但仍然面临此错误。

卷曲位于这里

ᴾᴋᴹɴ Master Red ▰ ◓ ◓ ◓ ◓ ◓ ◓ 

> curl-config --cflags
-I/usr/local/Cellar/curl/7.64.1/include

卷曲图书馆包括

ᴾᴋᴹɴ Master Red ▰ ◓ ◓ ◓ ◓ ◓ ◓ 

> curl-config --libs
-L/usr/local/Cellar/curl/7.64.1/lib -lcurl -lldap -lz

即使添加到我的包含路径后，仍然会得到相同的错误。

# c++ curl support for the g++ compiler
# curl-config --cflags
CPLUS_INCLUDE_PATH="/usr/local/Cellar/curl/7.64.1/include:${CPLUS_INCLUDE_PATH}"
# curl-config --libs which include -lcurl -lldap -lz
CPLUS_INCLUDE_PATH="/usr/local/Cellar/curl/7.64.1/lib:${CPLUS_INCLUDE_PATH}"
export CPLUS_INCLUDE_PATH

---

错误

ᴾᴋᴹɴ Master Red ▰ ◓ ◓ ◓ ◓ ◓ ◓ 

> g++ -std=c++17 main.cpp && ./a.out
Undefined symbols for architecture x86_64:
  "_curl_easy_cleanup", referenced from:
      _main in main-9d2f7a.o
  "_curl_easy_init", referenced from:
      _main in main-9d2f7a.o
  "_curl_easy_perform", referenced from:
      _main in main-9d2f7a.o
  "_curl_easy_setopt", referenced from:
      _main in main-9d2f7a.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

---

资源

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

int main()
{
    CURL *curl = curl_easy_init();
    if(curl) {
        CURLcode res;
        curl_easy_setopt(curl, CURLOPT_URL, "http://example.com");
        res = curl_easy_perform(curl);
        curl_easy_cleanup(curl);
    }
}

Answer 1

使用makefile而不是在命令行上调用make ，你就可以了

# *****************************************************
# Variables to control Makefile operation

CXX = g++

# ****************************************************
# Targets needed to bring the executable up to date

main: main.cpp
    $(CXX) main.cpp -I/usr/local/Cellar/curl/7.64.1/include -L/usr/local/Cellar/curl/7.64.1/lib -lcurl -lldap -lz && ./a.out