C语言 字符数组转换整数 stringToInt

工作了一段时间，一直在公司平台上面码代码，有一天同事分享，突然发现自己连基本的知识都快忘记了

字符串转化z整数数组是考验程序员基本功的经典题目

int stringToInt(const char *str){}

先写一个简单的实现

int stringToInt(const char *str){char *p = str;int nNUM = 0;while (*p != '\0'){nNum = nNUM * 10 + (*p - '0');p++;}return nNUM;}

主要逻辑应该就是这样，每取一位字符，用字符和'0'比较计算数值，累加到前面的数值*10上面

 前面的基础上考虑一下正负数问题

int stringToInt(const char *str){char *p = str;int nNUM = 0;bool bFlage = true;if (*p == '-'){bFlage = false;p++;}else if (*p == '+'){p++;}while (*p != '\0'){nNum = nNUM * 10 + (*p - '0');p++;}if (!bFlage){nNUM = 0 - nNUM;}return nNUM;}

除了+-还有其他非法字符，需要容错处理

int stringToInt(const char *str){char *p = str;int nNUM = 0;bool bFlage = true;if (*p == '-'){bFlage = false;p++;}else if (*p == '+'){p++;}while (*p != '\0'){if ((*P < '0') || (*p > '9')){nNUM = 0;return nNUM;}nNum = nNUM * 10 + (*p - '0');p++;}if (!bFlage){nNUM = 0 - nNUM;}return nNUM;}

上面的代码无法判断0 是成功转换的字符串，还是非法数据之后的返回值，因此定义一个枚举常量，保存每次转换是否成功，可以使用系统枚举值，一般库函数都是自带的，保存对应操作是否成功

enum NumState{OK = 1,ERROR}int stringToInt(const char *str){char *p = str;int nNUM = 0;bool bFlage = true;enum NumState oState = OK;if (*p == '-'){bFlage = false;p++;}else if (*p == '+'){p++;}while (*p != '\0'){if ((*P < '0') || (*p > '9')){nNUM = ERROR;return nNUM;}nNum = nNUM * 10 + (*p - '0');p++;}if (!bFlage){nNUM = 0 - nNUM;}return nNUM;}

在考虑一下，除了上面问题应该还有溢出的问题，增加溢出的判断

int stringToInt(const char *str){char *p = str;int nNUM = 0;bool bFlage = true;enum NumState oState = OK;if (*p == '-'){bFlage = false;p++;}else if (*p == '+'){p++;}while (*p != '\0'){if ((*p < '0') || (*p > '9')){oState = ERROR;nNUM = 0;break;}if ((nNUM * 10)>0 && (*p - '0')&& (2147483647 - (nNUM * 10) < (*p - '0'))){oState = ERROR;nNUM = 0;break;}nNUM = nNUM * 10 + (*p - '0');p++;}if (!bFlage){nNUM = 0 - nNUM;}return nNUM;}

测试一下：

#include <stdio.h>#include <stdbool.h>enum NumState{OK = 1,ERROR};int stringToInt(const char *str){char *p = str;int nNUM = 0;bool bFlage = true;enum NumState oState = OK;if (*p == '-'){bFlage = false;p++;}else if (*p == '+'){p++;}while (*p != '\0'){if ((*p < '0') || (*p > '9')){oState = ERROR;nNUM = 0;break;}if ((nNUM * 10)>0 && (*p - '0')>0&& (2147483647 - (nNUM * 10) < (*p - '0'))){oState = ERROR;nNUM = 0;break;}nNUM = nNUM * 10 + (*p - '0');p++;}if (!bFlage){nNUM = 0 - nNUM;}return nNUM;}int main(){int nNum;nNum = stringToInt("-2147483647");printf("%d\n", nNum);return 0;}

不要忽视容错机制，很重要