atoi() 转换错误

发布于 2024-07-07 06:52:05 字数 331 浏览 9 评论 0原文

atoi() 给了我这个错误：


error C2664: 'atoi' : cannot convert parameter 1 from 'char' to 'const char *'
        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast

从这一行： int pid = atoi( token.at(0) ); 其中 token 是一个向量

我该如何解决这个问题？

原文

atoi() is giving me this error:


error C2664: 'atoi' : cannot convert parameter 1 from 'char' to 'const char *'
        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast

from this line:
int pid = atoi( token.at(0) );
where token is a vector

how can i go around this?

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丑疤怪 2024-07-14 06:52:05

token.at(0) 返回单个字符，但 atoi() 需要一个字符串（指向字符的指针）。要么将单个字符转换为字符串，要么将单个数字字符转换为它代表的数字通常可以^*这样做：

int pid = token.at(0) - '0';

^{* 例外情况是字符集没有按顺序编码数字 0-9，这种情况极为罕见。}

token.at(0) is returning a single char, but atoi() is expecting a string (a pointer to a char.) Either convert the single character to a string, or to convert a single digit char into the number it represents you can usually^* just do this:

int pid = token.at(0) - '0';

^{* The exception is when the charset doesn't encode digits 0-9 in order which is extremely rare.}

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遗心遗梦遗幸福 2024-07-14 06:52:05

您必须创建一个字符串：

int pid = atoi(std::string(1, token.at(0)).c_str());

... 假设 token 是 char 的 std::vector ，并使用接受单个字符的 std::string 构造函数（以及字符串将包含的该字符的数量，一个在这种情况下）。

You'll have to create a string:

int pid = atoi(std::string(1, token.at(0)).c_str());

... assuming that token is a std::vector of char, and using std::string's constructor that accepts a single character (and the number of that character that the string will contain, one in this case).

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GRAY°灰色天空 2024-07-14 06:52:05

您的示例不完整，因为您没有说出向量的确切类型。我假设它是 std::vector（也许，您用 C 字符串中的每个字符填充）。

我的解决方案是在 char * 上再次转换它，这将给出以下代码：

void doSomething(const std::vector & token)
{
    char c[2] = {token.at(0), 0} ;
    int pid   = std::atoi(c) ;
}

请注意，这是一个类似 C 的解决方案（即，在 C++ 代码中相当难看），但它仍然有效。

Your example is incomplete, as you don't say the exact type of the vector. I assume it is std::vector<char> (that, perhaps, you filled with each char from a C string).

My solution would be to convert it again on char *, which would give the following code:

void doSomething(const std::vector & token)
{
    char c[2] = {token.at(0), 0} ;
    int pid   = std::atoi(c) ;
}

Note that this is a C-like solution (i.e., quite ugly in C++ code), but it remains efficient.

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浅唱ヾ落雨殇 2024-07-14 06:52:05

const char tempChar = token.at(0);
int tempVal = atoi(&tempChar);

const char tempChar = token.at(0);
int tempVal = atoi(&tempChar);

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呆° 2024-07-14 06:52:05

stringstream ss;
ss << token.at(0);
int pid = -1;
ss >> pid;

例子：

#include <iostream>
#include <sstream>
#include <vector>

int main()
{
  using namespace std;

  vector<char> token(1, '8');

  stringstream ss;
  ss << token.at(0);
  int pid = -1;
  ss >> pid;
  if(!ss) {
    cerr << "error: can't convert to int '" << token.at(0) << "'" << endl; 
  }

  cout << pid << endl;
  return 0;
}

stringstream ss;
ss << token.at(0);
int pid = -1;
ss >> pid;

Example:

#include <iostream>
#include <sstream>
#include <vector>

int main()
{
  using namespace std;

  vector<char> token(1, '8');

  stringstream ss;
  ss << token.at(0);
  int pid = -1;
  ss >> pid;
  if(!ss) {
    cerr << "error: can't convert to int '" << token.at(0) << "'" << endl; 
  }

  cout << pid << endl;
  return 0;
}

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