istringstream十进制整数输入到8位类型 [英] istringstream decimal integer input to 8-bit type

问题描述

这：

  #include< iostream> 
 #include< sstream> 
＃include< inttypes.h> 
 
 using namespace std; 
 
 int main（void）{
 istringstream iss（123 42）; 
 int8_t x; 
 while（iss>> x）{
 cout<< x < endl; 
} 
 return 0; 
} 
  

产生：

  1 
 2 
 3 
 4 
 2 
  

但我想要：

  123 
 42 
  

铸造 iss>> （int）x （我最初尝试使用 char ）给我错误： （'istringstream'（aka'basic_istringstream'）and'int'）（clang）或错误：对'operator >>' b


是否有一种方法可以将值直接读取为 8位类型，或者必须使用中间商店吗？

解决方案

您必须使用中间类型或自己解析。所有char类型（char，signed char和unsigned char）都被视为文本元素，而不是整数。

注意：

• 输出会遇到相同的问题。


• 不要使用C风格的转换，它们几乎只会导致错误。


• 在输入操作之前检查EOF是无用的，您之后需要检查是否失败。

This:

#include <iostream>
#include <sstream>
#include <inttypes.h>

using namespace std;

int main (void) {
    istringstream iss("123 42");
    int8_t x;
    while (iss >> x) {
        cout << x << endl;
    }
    return 0;
}  

Produces:

1
2
3
4
2

But I want:

123
42

Casting iss >> (int)x (I initially tried this with a char) gives me "error: invalid operands to binary expression ('istringstream' (aka 'basic_istringstream') and 'int')" (clang) or "error: ambiguous overload for ‘operator>>’" (g++).

Is there a way to read the value as a number directly into an 8-bit type, or do I have to use an intermediary store?

解决方案

You have to use an intermediate type or do the parsing yourself. All char-types (char, signed char and unsigned char) are treated as text elements, not integers. int8_t is probably just a typedef for one of them, which is why your code fails.

Note:

• The output will suffer from the same issues.
• Don't use C-style casts, they almost only cause errors.
• Checking for EOF before an input operation is useless, you need to check for failure afterwards instead.

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