itoa函數實現

//整形轉成字串函數實現//題目不難，重點考察面試者對問題考慮的全面程度#include <iostream>using namespace std;void itoa_mf(int num,char str[]){int sign = num;int i = 0;int j = 0;char temp[100];//如果是負數就去掉符號,將-1234轉成1234if(sign < 0){num = -num;}//轉成字串，1234轉成"4321"do{temp[i] = num % 10 + '0';num /= 10;i++;}while(num > 0);//如果是負數的話，加個符號在末尾，如："4321-"if(sign < 0){temp[i++] = '-';}temp[i] = '\0';i--;//將temp數組中逆序輸入到str數組中//將"4321-" ====> "-1234"while(i >= 0){str[j] = temp[i];j++;i--;}//字串結束標識str[j] = '\0';}//測試案例void main(){int a = 0;char s[100];itoa_mf(a,s);cout << s << endl;}