Question

请帮助我如何解决这个问题......

#include <iostream>
#include <stdio.h>
#include <stdlib.h>

int main(int iics, char*argv[]){
int a,n,sum=0,average=0,highest=0,lowest=0,num[100];

for (a=1;a<iics;++a)
sum= sum + atoi (argv[a]);
printf ("The sum is %.2d \n", sum);
average=sum/a;
printf ("The average is %.2d \n", average);

highest=num[0];
lowest=num[0];
for (a=1;a<num[100];++a){
    if (num[a]>highest)
    highest=num[a];
    if(num[a]<lowest)
    lowest=num[a];}
    printf ("The highest element is %.2d \n",highest);
    printf ("The lowest element is %.2d \n", lowest);

return 0;

}

我得到了总和和平均值，但最高和最低的是怎样的？

Answer 1

您正在使用未初始化的数组num。所以这样做

int i=0;
for (a=1;a<iics;++a)
{
sum= sum + atoi (argv[a]);
num[i++]=atoi9argv[a]);// just for an idea. Take care about i value

}

自己尝试......

Answer 2

您希望从1到iics-1枚举您的其他循环，而不是从1枚举到num[a]-1

变化：

for (a=1;a<num[100];++a){

为：

for (a=1;a<iics;a++){

在您之前的循环中，将atoi (argv[a])分配给num[a]。然后这些将需要从num[1]开始：

highest=num[1];
lowest=num[1];

Answer 3

#include <stdio.h>
#include <stdlib.h>

int main(int iics, char*argv[]){
    int a,n=0,sum=0,average,highest,lowest,num[100];

    for (a=1; a<iics; ++a)
        sum = sum + (num[n++] = atoi(argv[a]));
    printf ("The sum is %.2d \n", sum);
    average = sum / n;
    printf ("The average is %.2d \n", average);

    highest = num[0];
    lowest = num[0];
    for (a = 1; a < n; ++a){
        if(num[a] > highest)
            highest = num[a];
        if(num[a] < lowest)
            lowest = num[a];
    }
    printf ("The highest element is %.2d \n",highest);
    printf ("The lowest element is %.2d \n", lowest);

    return 0;
}

如何使用参数和数组找到C中最大和最小的元素？

3 个答案: