c - 对与c中的字符串数组相关的数组进行排序

Question

我试图根据数组的值对与数组相关的字符串进行排序，由于某种原因，排序部分开始工作..当我尝试对彼此相关的元素进行排序并出于某种原因打印它们时打印结果非常随机

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int sorting(char name[][10],double average[],int size);

int main()
{
    double sales1[10],sales2[10],sales3[10],average[10],test,totalm=0,totalfm=0,test2;          
int i=0,j;
char name[10][10],gender[10];
printf("Please input the name, the gender, and the sales for the\nfirst three months followed by a spacebar after each element\n");
scanf("%s %c %lf %lf %lf",&name[0],&gender[0],&sales1[0],&sales2[0],&sales3[0]);
average[i]=(sales1[i]+sales2[i]+sales3[i])/3;
while(strcmp(name[i],"enough")!=0)
{
     i++;
     printf("Please input the name, the gender, and the sales for the\nfirst three months followed by a spacebar after each element\n");
     scanf("%s %c %lf %lf %lf",&name[i],&gender[i],&sales1[i],&sales2[i],&sales3[i]);
     average[i]=(sales1[i]+sales2[i]+sales3[i])/3;
}
sorting(name,average,i);
j=i;
while(i>=0)
 {
    if(gender[i]=='m')
      totalm=totalm+average[i];
    else
      totalfm=totalfm+average[i];
    i--;
 }
   while(j>=0)
    {
    test2=strcmp(name[j],"enough");
    if(test2!=0)
      printf("%s\t%f\n",name[j],average[j]);
      j--;
      }
      printf("total male sales are %f\n",totalm);
      printf("total female sales are %f\n",totalfm);


  }
 int sorting(char name[][10],double average[], int size)
 {
int i=0;
double temp;
char ntemp[20][20];
while(i<=size)
 {
     if(average[i+1]>average[i])
      {
          temp=average[i];
          strcpy(ntemp[i],name[i]);
          average[i]=average[i+1];
          strcpy(name[i],name[i+1]);
          average[i+1]=temp;
          strcpy(name[i+1],ntemp[i]);
      }
    i++;
 }
}

谢谢!

Answer 1

我认为您应用的排序在排序函数的第一个条件上似乎是错误的

while(i<=size)
 {
     if(average[i+1]>average[i])
      {
          temp=average[i];

假设当 i 等于 size 时，平均值 [i+1] 将不指向任何内容，您可以说该值为零，但您没有设置该值。所以尝试更正此代码

for(i=0;i<n;i++)
  {
   for(k=0;k<n-i-1;k++)
    {
     if(a[k]>a[k+1])
      {
       temp=a[k];
       a[k]=a[k+1];
       a[k+1]=temp;
      }
    }
  }

这是冒泡排序，您总是比最后一次迭代少一次。了解更多see here