无论用户输入如何,将输入转换为int

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

int main()
{
  int i, times, total;
  float average = 0.0;
  int * pArr;

  printf("For how many numbers do you want the average calculated?\n");

  scanf("%d", &times);

  pArr = (int *) malloc(times * sizeof(int));

  printf("Please enter them down here\n");

  for(i=0;i<times;i++) {
    scanf("%d", &pArr[i]);
    total += pArr[i];
  }

  average = (float) total / (float) times;
  printf("The average of your numbers is %.2f\n", average);

  return 0;
}

所以现在的问题是,当用户输入一个浮点数时,程序就会终止.任何线索？

但您可以通过扫描字符串,然后扫描字符串中的整数来解决它.这是错误的错误检查,但至少它会丢弃浮动部分,你可以输入浮点数或整数(还要注意总数没有初始化,gcc -Wall -Wextra甚至没有检测到).

找到下面的工作版本(虽然输入整数时需要更多的错误检查)：

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

int main()
{
  int i, times, total = 0;  // don't forget to initialize total to 0 !!!
  float average = 0.0;
  int * pArr;
  char temp[30];
  printf("For how many numbers do you want the average calculated?\n");

  scanf("%d", &times);

  pArr = malloc(times * sizeof(int)); // don't cast the return value of malloc

  printf("Please enter them down here\n");

  for(i=0;i<times;i++) {
    scanf("%29s",temp);   // scan in a string first
    sscanf(temp,"%d", &pArr[i]);  // now scan the string for integer
    total += pArr[i];
  }

  average = (float) total / (float) times;
  printf("The average of your numbers is %.2f\n", average);

  free(pArr);  // it's better to free your memory when array isn't used anymore
  return 0;
}

笔记：

>阵列分配&如果你只计算值的平均值,那么存储在这里没用>你没有受到有效浮点指数输入的保护：如果输入1e10,则扫描值1(与另一个可行的答案的“％f”方法相反,但我担心存在风险舍入错误)