c++如何输出argc与argv

Question

为什么argc用printf可以打印出来，用cout就不可以？

#include<vector>
#include <iostream>
#include <stdio.h>
int main(int argc,char** argv)
{
    int con=0;
    std::cout<<con<<std::endl;
    std::cout<<"c++ "<<agrc<<std::endl;
    printf("argc= %d",argc);

}

a.cpp: In function ‘int main(int, char**)’:
a.cpp:8:24: error: ‘agrc’ was not declared in this scope
std::cout<<"c++ "<<agrc<<std::endl;

Answer 1

参数名称敲错了是argc，不是agrc
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