这个程序的错误是什么？它显示了一般保护故障错误 [英] what is the mistake in this program? it shows general protection fault error

问题描述

#include<stdio.h>
void main()
{

    char name[30];
    int i;

    printf("Enter ur name:\n");
    for(i=0;i<30;i++)
    scanf("%s",&name[i]);



    for(i=0;i<30;i++)
    printf("The name is :%s",name[i]);
}

推荐答案

你的printf语句是使用％s，用于ASCIIZ字符串，你打印的是单个字符name [i] - 被视为指向谁知道你的O / S在哪里字符串的指针



printf（％s， name）;

或printf（％c，name [i]）; //为你的循环

You're printf statement is using %s, which is for ASCIIZ strings and you are printing individual characters name[i] - which are being treated as pointers to a string in the "who knows where" of your O/S

either printf("%s", name);
or printf("%c", name[i]); // for your loop

// wf-protection-fault-error.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<stdio.h>

const int max_namecount   =5;   // 5 names
const int max_namelength  =50;  // each name has space for 50 chars

int main(int argc, char *argv[]) 
{
    char name[max_namecount][max_namelength+3];
    int i;

    printf("Enter your name:\n");
    for(i=0;i<max_namecount;i++)  scanf("%50s",name[i]); // 50=max_namelength:  makes sure, that you do not read more chars than you have space in name

    getc(stdin); // only for pause
    for(i=0;i<max_namecount;i++)  printf("\nThe name is :%s",name[i]);

    getc(stdin); // only for pause
      return 0;
}

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