求助：关于 int main(int argc, char *argv[]) 的问题

[ysysbaobei]

要求：
编写一个主函数main,使用两个值作为实参，并输出它们的和

[太乙]

#include<stdlib.h>
#include<stdio.h>
void main(int argc,char* args[])
{
    if(argc<3)
         return;
    printf("%d",atoi(args[1])+atoi(args[2]));
}

[freshairfly]

#include <iostream>
using namespace std;

 int main(int argc, char *argv[])
{
    if (argc != 3)
    {
        cout << "wrong parameter!" << endl;
        return 1;
    }

    int arg1 = atoi(argv[1]);
    int arg2 = atoi(argv[2]);

    cout << "result=" << arg1 + arg2 << endl;

    return 0;
}

[darkwalkertt]

#include<iostream.h>
int main(int argc,char *argv[]){
if(argc!=3)return 0;
else{
int num1=atoi(argv[1]);
int num2=atoi(argv[2]);
cout<<"result="<<num1+num2<<endl;
return 1;
}
}

[xhs_lh04]

楼上几位都说了,没啥补充的,UP

[jia_xiaoxin]

#include <iostream>
using namespace std;

int main(int argc, char *argv[])
{
    if (argc != 3)
    {
        return 1;
    }
    int result = itoa(argv[1]) + itoa(argv[2]);
    cout << result << endl;
}

[ysysbaobei]

问题已解决，谢谢大家

#include <iostream>
using namespace std;

int main(int argc, char *argv[])
{
	if (argc < 3)
		return 0;
	int sum(0);
	for (int i = 1; i < argc; ++i)
		sum += atoi(argv[i]);
	cout << "sum is : " << sum << endl;

	return 0;
}

[ysysbaobei]

结贴了