C++位移在一个参数

Question

嘿家伙我的任务基本上是从char获得小数。例如，输入参数是8751.我怎样才能得到51（只）转换为十进制？我知道如何得到87，我也知道如果得到51如果它是像十六进制（0x8751 & 0xff）。

所以我的程序的输出将是：

(0x87 & 0x7f)*128 + (0x51) 

OR

(135-128)*128 + 81 

感谢

代码::

#include <iostream> 
#include <cstdio> 


int 
main(int argc, char* argv[]) 
{ 
char* nums; 
// long sum = 0; 
for (int i = 1; i < argc; ++i) 
{ 
    nums=argv[i]; 
// long a; 
// ::sscanf(argv[i], "%lx", &a); // read hex string and form long value 
// sum += a; // same as 'sum = sum + a' 
} 
int len = strlen(nums); 

if(len<=2){ 
    unsigned long cur; 
    ::sscanf(nums, "%lx", &cur); 
    if(cur<=128){ 
     std::cout<<cur; 
    } 
} 

    if(len=4){ 
    unsigned long cur; 
    ::sscanf(nums, "%lx", &cur); 
    char tr []= "0x"; 
    //std:: cout << cur<<"\n"; 
    unsigned long cur1 = nums & 0xff; 
    unsigned long cur2 = cur >> 8; 
    if(cur1 >128){ 
     //std :: cout <<cur1<<"\n"; 
     std::cout<<0; 
    } 
    else{ 
     unsigned long result = (cur2 - 128)*128 + cur1; 
      std ::cout << result; 
    } 
    } 


    system("pause"); 
    return 0; 
} 

Answer 1

是输入总是4个字符长？如果是这样的......

#include <stdio.h> 

int main(int argc, char *argv[]) { 
    char num [] = "8751"; 
    int hi, lo; 
    sscanf(num, "%2x%2x", &hi, &lo); 
    printf("%s -> %d, %d\n", num, hi, lo); 
} 

打印8751 -> 135, 81。如果num被更改，您将获得81ab -> 129, 171或a1b7 -> 161, 183。