在scanf之后使用fgets [英] Using fgets after scanf

问题描述

我想摆脱缓冲区溢出，我使用了 fgets 而不是 scanf 来限制字符.但是，每当我在 fgets 之前用 scanf 输入内容时，它就无法正常工作.

I want to get rid of buffer overflow and I am using fgets instead of scanf to limit the characters. But whenever I enter something with scanf before fgets, it doesn't work correctly anymore.

此代码正常工作

#include <stdio.h>
int main()
{
 char name[10];
 printf("Who are you? \n");

 fgets(name,10,stdin);                   

 printf("Good to meet you, %s.\n",name);
 return(0);
}

此代码无法正确读取名称

This code does not read name correctly

#include <stdio.h>
#include <stdlib.h>
#define MAX 15

int new_acc();
int view_list();
int main(){    
    int one=1, two=2, three=3, four=4, five=5, six=6, seven=7, choice;
    int new_account, list;

    printf("%d. Create new account\n",one);

    printf("Enter you choice: ");
    scanf("%d",&choice);

    if (choice==one){new_account = new_acc();}
    return 0;
}

int new_acc(){
    char name[MAX], address, account;

    printf("Enter your name: ");
    fgets(name, MAX, stdin);          /* it is the code */
}

为什么会发生，我该如何解决?

Why is it happening and how do I fix it?

推荐答案

请勿在同一代码中将 scanf()与 fgets()混合使用.

Don't mix scanf() with fgets() in the same code.

scanf(％d"，& choice); 不会消耗掉所有的 line -它留下了结尾的'\ n'，以便以后的 fgets()用作空行.

scanf("%d",&choice); does not consume all the line - it leaves the trailing '\n' for the later fgets() to consume as an empty line.

scanf()很难以安全的方式使用.

scanf() is difficult to use in a secure fashion.

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