如何将任何用户输入都神奇地转换成整数呢？

本文共计548个文字，预计阅读时间需要3分钟。

c#include

int main() { float input; printf(请输入一个浮点数：); scanf(%f, &input); int output=(int)input; printf(转换后的整数为：%d\n, output); return 0;}

我有以下c代码,并希望确保即使用户输入一个浮点数,这将转换为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？

scanf遇到点时会停止扫描.因此无法直接扫描输入.

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)