本文介绍了C ++ random 0xC0000005错误的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 29岁程序员,3月因学历无情被辞! 我制作了一个将 n 十进制数字 sk 转换为其他数字系统 p 的程序,但有时它会崩溃,我得到的错误代码是0xC0000005(程序仍然转换并输出所有数字)。一个事情我只是注意到它发生,然后转换数字超过6个符号(或者只是一个巧合)。 #include < iostream> #include< fstream> #include< string> using namespace std; int main() { long n,sk,p,j; string liekanos; ifstream f(u1.txt); f>> n; for(int i = 0; i { f> sk>> p; j = 0; while(sk> 0) { liekanos [j] = sk%p; sk / = p; j ++; } for(j> = 0; j--;) { if(liekanos [j] cout else cout<< char(liekanos [j] +55); } cout<< endl; } return 0; } 输入示例: 3 976421618 7 15835 24 2147483647 2 $ b 使用 liekanos [j] 您访问索引时的元素,但由于你没有指定这个字符串的大小,你很可能试图访问非现有的元素。你可以在输入和之前调用 liekanos.resize(sk),以确保不会发生。 或者,如果你知道 liekanos 的最大大小,你可以声明为 string liekanos (N,c); 其中 N 是其大小, c 是默认值的每个字符。 I made a program which converts n decimal numbers sk into other numerical system p but sometimes it crashes and the error code I get is 0xC0000005 (program still converts and outputs all the numbers) . One thing I just noticed that it happens then converted number is longer than 6 symbols (or it's just a coincidence).#include <iostream>#include <fstream>#include <string>using namespace std;int main(){ long n,sk,p,j; string liekanos; ifstream f("u1.txt"); f >> n; for (int i=0;i<n;i++) { f >> sk >> p; j=0; while (sk>0) { liekanos[j]=sk % p; sk/=p; j++; } for (j>=0;j--;) { if (liekanos[j]<10) cout<<int(liekanos[j]); else cout<<char(liekanos[j]+55); } cout<<endl; } return 0;}Example input:3976421618 715835 242147483647 2 解决方案 With liekanos[j] you access element at index j but since you haven't specify size of this string, you are most likely trying to access non-existing element. You could call liekanos.resize(sk) before you enter your while loop to make sure it never happens.Or if you know the maximum possible size of liekanos, you could declare it as string liekanos(N, c); where N is its size and c is the default value of each character in it. 这篇关于C ++ random 0xC0000005错误的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云! 06-15 03:20