强制字符串到int消耗整个字符串的函数 [英] Forcing String to int Function to Consume Entire String

问题描述

给定一个应该代表一个数字的字符串，我想把它放到一个转换函数中，如果整个字符串没有转换，它会提供通知。

输入：12：

• code> istringstream :: operator>> 输出12


• li>

• stoi 输出12



1X我想要一个失败响应，但我得到：



• code> istringstream :: operator>> 输出1


• li>

• stoi 输出1



X2：

• istringstream :: operator> > 输出0并设置错误标志


• atoi b
• stoi 发生错误

[ Live Example ]

有一种方法可以引起输入的错误行为1X？

解决方案

com / w / cpp / string / byte / strtolrel =nofollow> strtol

正如建议的这里 strtol 的out-parameter可以使用以获取读取的字符数。 strtol 实际上会返回 long 而不是 int

  char * size; 
 const int num = strtol（i.c_str（），& size，10）; 
 
 if（distance（i.c_str（），const_cast< const char *>（size））== i.size（））{
 cout< strtol：< num<< endl; 
} else {
 cout<< strtol：error\\\
; 
} 
  

注意，这使用 i.c_str / code>来引用同一个字符串。 c_str 退款如果你有C ++ 11，指向底层数组的指针不是临时的：



）和 data（）执行相同的函数

另请注意， c_str 返回的指针将在 strtol 和 distance之间有效除非：

• 传递非 - const c>引用任何标准库函数的字符串



• 调用非 - <$ c $在字符串上除了运算符[] ，<$ c> $ c> at（）， front（）， back（）， begin（）， rbegin（）， code> rend（）

在这些情况下，您需要创建 i 的基础 const char * 的临时副本，然后执行测试

sscanf

sscanf 可以使用％n 返回读取的字符数，这可能比执行指针比较更直观。如果base很重要， sscanf 可能不是一个好的选择。与 strtol 和 stoi 不同，支持基础2 - 36， sscanf 提供了八进制（％o ），十进制（％d ）和十六进制（％x ）。

  size_t size; 
 int num; 
 
 if（sscanf（i.c_str（），％d％n，& num，& size）== 1&& 
 cout<< sscanf：< num<< endl; 
} else {
 cout<< sscanf：error\\\
; 
} 
  

stoi

此处 stoi 的输出参数像 sscanf 的的工作原理％n 读取的字符数。与C ++相比，这需要一个 string ，而不像上面的C实现 stoi throws a invalid_argument 如果未考虑第一个非空格字符这不幸意味着不像C的实现，这必须检查在 try 和 catch 阻止。

  try {
 size_t size; 
 const auto num = stoi（i，& size）; 
 
 if（size == i.size（））{
 cout< stoi：< num<< endl; 
} else {
 throw invalid_argument（invalid stoi argument）; 
} 
} catch（const invalid_argument& / * e * /）{
 cout< stoi：error\\\
; 
} 
  

[ Live Example ]

Given a string that should represent a number, I'd like to put it into a conversion function which would provide notification if the whole string did not convert.

For input: "12":

• istringstream::operator>> outputs 12
• atoi outputs 12
• stoi outputs 12

For input "1X" I'd like a failure response but I get:

• istringstream::operator>> outputs 1
• atoi outputs 1
• stoi outputs 1

For input "X2":

• istringstream::operator>> outputs 0 and sets an error flag
• atoi outputs 0
• stoi throws an error

[Live Example]

Is there a way to provoke the error behavior on input "1X"?

解决方案

For a given string str there are several ways to accomplish this each with advantages and disadvantages.

strtol

As suggested here strtol's out-parameter can be used to get the number of characters read. strtol actually returns a long not an int so a cast is happening on the return.

char* size;
const int num = strtol(i.c_str(), &size, 10);

if(distance(i.c_str(), const_cast<const char*>(size)) == i.size()) {
    cout << "strtol: " << num << endl;
} else {
    cout << "strtol: error\n";
}

Note that this uses i.c_str() to refer to the same string. c_str Returns pointer to the underlying array serving as character storage not a temporary if you have C++11:

c_str() and data() perform the same function

Also note that the pointer returned by c_str will be valid between the strtol and distance calls unless:

• Passing a non-const reference to the string to any standard library function
  
• Calling non-const member functions on the string, excluding operator[], at(), front(), back(), begin(), rbegin(), end() and rend()

If you violate either of these cases you'll need to make a temporary copy of i's underlying const char* and perform the test on that.

sscanf

sscanf can use %n to return the number of characters read which may be more intuitive than doing a pointer comparison. If base is important, sscanf may not be a good choice. Unlike strtol and stoi which support bases 2 - 36, sscanf provides specifiers for only octal (%o), decimal (%d), and hexadecimal (%x).

size_t size;
int num;

if(sscanf(i.c_str(), "%d%n", &num, &size) == 1 && size == i.size()) {
    cout << "sscanf: " << num << endl;
} else {
    cout << "sscanf: error\n";
}

stoi

As suggested here stoi's output parameter works like sscanf's %n returning the number of characters read. In keeping with C++ this takes a string and unlike the C implementations above stoi throws an invalid_argument if the first non-whitespace character is not considered a digit for the current base, and this unfortunately means that unlike the C implementations this must check for an error in both the try and catch blocks.

try {
    size_t size;
    const auto num = stoi(i, &size);

    if(size == i.size()) {
        cout << "stoi: " << num << endl;
    } else {
        throw invalid_argument("invalid stoi argument");
    }
} catch(const invalid_argument& /*e*/) {
    cout << "stoi: error\n";
}

[Live Example]

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