错误：不能将'std :: basic_string< char> :: iterator ...'转换为'const char *'参数'1'...' [英] error: cannot convert 'std::basic_string<char>::iterator ...' to 'const char* for argument '1' ...'

问题描述

我收到以下错误：

error: cannot convert 'std::basic_string<char>::iterator {aka __gnu_cxx::__normal
_iterator<char*, std::basic_string<char> >}' to 'const char*' for argument '1' 
to 'int remove(const char*)'

由于某种原因，我的程序编译完美，当我在Mac上工作...但一旦我使用Linux机器，这个错误弹出在多个地方。 （如果任何人可以解释为什么会发生这种情况，那就太棒了！）

For some reason, my program compiles perfectly when I'm working on a Mac... but once I use a Linux machine, this error pops up in more than one place. (If anyone could explain why this occurs, that would be great!)

这里是其中一个错误弹出的实例：

Here's one of the instances where the error pops up:

SomeClass::SomeClass(string t, string art, Time dur) {
    char chars[] = ",";
    t.erase(std::remove(t.begin(), t.end(), chars[0]), t.end());
    art.erase(std::remove(art.begin(), art.end(), chars[0]), art.end());
    // Some more code ...
}

错误来自此行：

t.erase(std::remove(t.begin(), t.end(), chars[0]), t.end());

有没有人知道如何解决这个问题？

Does anyone know how to approach this problem?

推荐答案

您忘记了 #include< algorithm> 找到了一个href =http://en.cppreference.com/w/cpp/algorithm/remove> std :: remove 。否则，您的编译器只知道此 std :: remove

You forgot to #include <algorithm>, where std::remove is located. Without that, your compiler only knows about this std::remove (I get the same error with Visual C++ 14), which is defined in indirectly included <cstdio> header.

编译器中不同的行为是标准库实现的不同 #include 层次结果。

Different behavior among compilers is a result of different #include hierarchies of the standard library implementations.

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