getline没有匹配功能

Question

当我只想在blankspace之前读取字符时,使用:' cin >> buffer.但是,我想读取用户输入的所有内容,包括空格.所以,我从使用">>"更改为getline调用.我似乎传递了正确的参数,我已经完成了#include和#include.编译时的错误消息是:

Driver.cpp:在函数'int main(int,char*const*)'中:Driver.cpp:49:44:错误:没有用于调用'getline(std :: istream&,char [1024])'getline的匹配函数(cin,buffer); ^ Driver.cpp:49:44:注意:候选者是:/usr/include/wchar.h:4:0中包含的文件,来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c ++/cwchar:44,来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/bits/postypes.h:40,来自/ usr/lib/gcc/x86_64-pc-cygwin来自/ usr/lib/gcc/x86_64 -cygwin/4.8.2/include/c ++/ostream:38,来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/iostream:39,来自Driver.cpp:1:/ usr /include/sys/stdio.h:37:9:注意:ssize_t getline(char**,size_t*,FILE*)ssize_t _EXFUN(getline,(char**,size_t*,FILE*)); ^ /usr/include/sys/stdio.h:37:9:注意:候选人需要3个参数,2提供来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/的文件字符串:52:0,来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/bits/locale_classes.h:40,来自/ usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c ++/bits/ios_base.h:41,来自/ usr/lib/gcc /来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/ios:42 x86_64-pc-cygwin/4.8.2/include/c ++/ostream:38,来自/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/iostream:39,来自Driver.cpp:1 :/usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/bits/basic_string.h:2793:5:注意:模板std :: basic_istream <_CharT,_Traits>&std :: getline( std :: basic_istream <_CharT,_Traits>&,std :: basic_string <_CharT,_Traits,_Alloc> &&)getline(basic_istream <_CharT,_Traits>&__ is,^

 ^ /usr/lib/gcc/x86_64-pc-cygwin/4.8.2/include/c++/bits/basic_string.tcc:1068:5:

注意:模板参数推断/替换失败:Driver.cpp:49:44:注意:不匹配的类型'std :: basic_string <_CharT,_Traits,_Alloc>'和'char [1024]'getline(cin,buffer); ^ Makefile:24:目标'Driver.o'的配方失败make:* [Driver.o]错误1

以下是删除了不属于此案例的代码的代码:

#include <iostream>
#include <cstdio>
#include <string>
#include <getopt.h>
#include "Driver.hpp"
#include "SymTab.hpp"

using namespace std;

#ifdef NULL
#undef NULL
#define NULL 0
#endif

ostream & operator << (ostream & stream, const Student & stu) {
        return stream << "name:  " << stu.name
                << " with studentnum:  " << stu.studentnum;
}

int main (int argc, char * const * argv) {
        char buffer[BUFSIZ];
        char command;
        long number;
        char option;

        while ((option = getopt (argc, argv, "x")) != EOF) {

        switch (option) {
                case 'x': SymTab<Student>::Set_Debug_On ();
                        break;
                }       
        }

        SymTab<Student> ST;
        ST.Write (cout << "Initial Symbol Table:\n" );

        while (cin) {
                command = NULL;         // reset command each time in loop
                cout << "Please enter a command ((i)nsert, "
                        << "(l)ookup, (r)emove, (w)rite):  ";
                cin >> command;

                switch (command) {

                case 'i': {
                        cout << "Please enter student name to insert:  ";
                        getline(cin, buffer);

                        cout << "Please enter student number:  ";
                        cin >> number;

                        Student stu (buffer, number);

                        // create student and place in symbol table
                        ST.Insert (stu);
                        break;
                }

        }

        ST.Write (cout << "\nFinal Symbol Table:\n");
}

这是我的第一篇文章,请告诉我是否搞砸了我的格式,如果需要更多信息来帮助我.谢谢!

Answer 1

std::getline()a (basic) istream和a的所有重载都是string.要解决问题,请char buffer []转到std::string buffer并记住#include <string>.