将字符串扫描为十六进制字符数组 [英] scanning a string to hex char array

问题描述

这是我的示例代码:

char a;
char str [20];
unsigned char b[8] ;
unsigned char c[8];

int argsread;
int i;

init_8051();

while(1)
{
    printf("
 enter a 64 bit operation 
");

    argsread = scanf("%s", str);

    sscanf(str, "0x%x%c0x%x", b, &a, c);

    printf("
 %d arguments read 
",argsread);

        for(i=0;i<8;i++)
{
             printf("
 %#x %c %#x
",b[i],a,c[i]);
}
        }

}

这里的问题是，当我在终端中输入例如以下输入时:0x1234567890abcdef+0x1234567890abcdef

The problem here is that when i enter for example in the terminal the following input : 0x1234567890abcdef+0x1234567890abcdef

这会导致错误，其中 char 数组的输出完全错误，并且大部分数字都被跟踪到 a 中，这应该是加号，我在做一些根本错误的事情吗?

this leads to an error where output for char array is totally wrong and most of the numbers are traced into a , which should have been the plus sign , am ai doing something fundamentally wrong ?

更新:我将代码更改为以下内容:

Update: I changed my code to the following :

while(1)
{
printf("
 enter a 64 bit operation 
");
argsread = scanf("%s", str);
sscanf(str, "0x%s%c0x%s", b, &a, c);
printf("
 %d arguments read 
",argsread);

printf("
 %s 
",b);
}

并且我将 str 的大小增加到 30 问题是程序的输出是:

and i increased size of str to 30 the problem is the output of the program is :

 1 arguments read 

 abcdef+0xabcdef 

所以 b 的值而不是只是 abcdef 整个字符串！

so the value of b instead of being just abcdef it the the whole string!!

更新 2:发现这段代码很完美，但我想使用 scanf 而不是 cin 这里是代码

Update2: Found this code which works perfect but i wanted to use scanf instead of cin here is the code

:`#include <iostream> 

using namespace std;

int main()
{

  float a, b, result;
    char oper, clear;
    cout << "Please enter an equation: i.e 5+5 " << endl;
    for (;;) {
          cin >> a;
          cin >> oper;
          cin >> b;
         if (oper == '+')
        result = a + b;
         else if (oper == '-')
            result = a - b;
       else if (oper == '*')
              result = a * b;
     else if (oper == '/')
              result = a / b;   
    cout << "= " << result << endl << endl;
    cin.clear();
    cin.ignore();
   } 
} `

推荐答案

如果代码以 fgets() 开头，则用户输入和错误处理会更容易.
然后使用sscanf()、strtol()等进行解析.

User input and error handling much easier if code starts with fgets().
Then use sscanf(), strtol(), etc. to parse.

printf("
 enter a 64 bit operation 
");
char buf[100];
if (fgets(buf, sizeof buf, stdin) == NULL) Handle_IOErrororEOF();

char a;
char b[17];  // use right size arrays
char c[17];
// use width of 16 and %[]
if (sscanf(buf, " 0x%16[0-9abcdef] %c 0x%16[0-9abcdef]", b, &a, c) != 3) {
  Handle_Bad_Input();
}

OTOH，只需使用允许十六进制输入 "%x" 或 "%i"

OTOH, just use an integer format specifier that allows hex input "%x" or "%i"

unsigned long long b,c;
if (sscanf(buf, "%lli %c%lli", &b, &a, &c) != 3) {
  Handle_Bad_Input();
}

<小时>

为什么 char str [20];scanf("%s", str);有问题:

"%s" 做了三件事:
1) 扫描但不保存所有前面的(0 个或多个)空格(' ', ' ', ' ' 等).
2) 扫描并保存所有非空白.
3)最终达到一个空白.它停止扫描并将空白放回 stdin.

"%s" does 3 things:
1) scans, but does not save, all preceding (0 or more) white-space (' ', ' ', ' ', etc.).
2) scans and saves all non-white-space.
3) Finally reaching a white-space. it stops scanning and puts that white-space back into stdin.

"%s" 说明符缺少宽度，如 "%19s"，因此很容易溢出 str

The "%s" specifier lacks a width, like "%19s", so it can easily overfill str

sscanf(str, "0x%s%c0x%s", b, &a, c); 也有问题.

输入在第一个数字的末尾和 '+' 之间没有空格，所以 "%s" 继续扫描.代码不检查 sscanf() 的返回值，然后使用 a、b、c.因此 a、b、c 可能无法正确扫描或初始化 - 导致潜在的未定义行为.

Input has no white-space between the end of the first number and the '+', so "%s" continues scanning. Code does not check the return value from sscanf() and then uses a, b, c. So a, b, c may not be properly scanned nor initialized - leading to potential undefined behavior .

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