SOJ.Hex to Int

最新推荐文章于 2019-12-23 13:25:11 发布

骚然勿外

最新推荐文章于 2019-12-23 13:25:11 发布

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分类专栏：刷题人生类相关文章标签： OJ 类 C++

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本文介绍了一个将十六进制字符串转换为十进制整数的C++函数实现，并展示了如何通过抛出异常来处理非法输入的情况。该函数能够处理正负数，并考虑了补码表示。

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Hex to Int


	时间限制：1秒内存限制：256兆题目描述 Write a function that parses a hex number as a string into a decimal integer. The function is defined as follows. int parseHex(const char * hexString) For example, hexString "A5" is 165 (1016+5=165). So, parseHex("A5") returns 165. Your implementation of parseHex function should throw a runtime_error exception if the binaryString is not a hex string. The main function is: int main() { string s; while (cin >> s) { try { cout << parseHex(s) << endl; } catch (runtime_error &ex) { cout << ex.what() << endl; } } return 0; } 样例输入 10 fffffffF 9G3 样例输出 16 -1 Hex number format error 提示 You also NEED to submit the main() function. The hex numbers mentioned above must be considered astwo's complements* of 32-bit numbers. (ffffffff)16, therefore, is less than 0, since its most significant bit is 1.


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#include
   
    
#include
    
     
#include 
     
      
#include
      
       
using namespace std;
int parseHex(const char * const hexString)
{
int result=0; //...
	int len = strlen(hexString);
	int dec = 0;
	for(int i=0; i
       
        ='0' && hexString[i]<='9')||(hexString[i]>='A' && hexString[i]<='F')||(hexString[i]>='a' && hexString[i]<='f'))
		{
		if(hexString[i]>='0' && hexString[i]<='9') dec=hexString[i]-'0';
		else if(hexString[i]>='A' && hexString[i]<='F') dec=hexString[i]-'A'+10;
		else if(hexString[i]>='a' && hexString[i]<='f') dec=hexString[i]-'a'+10;
		result=16*result+dec; 
		}
		else throw runtime_error("Hex number format error");
	}
return result;
}
int main()
{
string s;
while (cin>>s) {
try {
cout << parseHex(s.c_str()) << endl;
} catch (runtime_error &ex) {
cout << ex.what() << endl;
}
}
return 0;
}