c语言十个整数统计正整数的数量，要用函数调用

Question

如题所述

Answer 1

#include <stdio.h>
int func(int *arr,int n)
{
int i,count=0;
for(i=0;i<n;i++)
if(arr[i] > 0)
count++;
return count;
}
int main()
{
int a[10],i;
for(i=0;i<10;i++)
scanf("%d",&a[i]);
printf("正整数的数量为:%d个\n",func(a,10));
return 0;
}
示例运行结果：
-1 -2 4 5 6 -9 -8 10 -3 0
正整数的数量为:4个

Answer 2

#include<stdio.h>
void calc(){
int num[10],i,count=0;
for(i=0;i<10;i++){
printf("请输入第%d个数：",(i+1));
scanf("%d",&num[i]);
if(num[i]>0){
count++;
}
}
printf("正整数有%d个！\n",count);
}
void main(){
calc();
}

Answer 3

#include "stdio.h"

int positivenum(int a[],int n)
{
int i;
int k=0;
for(i=0;i<n;i++)
{
if(a[i]>0)
{
k++;
}
}
return k;
}
int main(int argc, char* argv[])
{
int a[10]={1,2,3,4,0,-6,-7,8,9,-20};
printf("positive = %d\n",positivenum(a,10));
return 0;
}