cout

问题 How would I capture the output from a php script that I run at the command line and store it to an array in c++? i.e. system('php api/getSome.php'); 回答1: Traditionally popen() is preferred when you need to IO between your parent and child process. It uses the C based FILE stream. It is a wrapper around the same low level intrinsics that also make up system(), namely fork + execl, but unlike system(), popen() opens a pipe and dups stdin/out to a stream for reading and writing. FILE *p = popen(

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: mixing cout and printf for faster output I'm using Microsoft Visual Studio 6.0. The following program, #include "stdafx.h" #include "iostream.h" int main(int argc, char* argv[]) { printf("a"); printf("b"); printf("c"); return 0; } produces "abc". While the following program, #include "stdafx.h" #include "iostream.h" int main(int argc, char* argv[]) { printf("a"); cout<<"b"; printf("c"); return 0; } produces "acb

问题 I would want to make a template of a << operator in C++, that would show a Object that is a "range" (by that i mean any object like : std::vector, std::set, std::map, std::deque). How can i achieve this? I've been googling and looking in docs for a few days now, but without any effect. I've been doing few templates and been overriding few operators before, but these were inside of a certain class that was representing a custom vector class. I cant seem to find a good way of implementing this,

问题 I would want to make a template of a << operator in C++, that would show a Object that is a "range" (by that i mean any object like : std::vector, std::set, std::map, std::deque). How can i achieve this? I've been googling and looking in docs for a few days now, but without any effect. I've been doing few templates and been overriding few operators before, but these were inside of a certain class that was representing a custom vector class. I cant seem to find a good way of implementing this,

问题 can't get "wcout" to print unicode string in multiple code pages, together with leaving "cout" to work please help me get these 3 lines to work together. std::wcout<<"abc "<<L'\u240d'<<" defg "<<L'א'<<" hijk"<<std::endl; std::cout<<"hello world from cout! \n"; std::wcout<<"hello world from wcout! \n"; output: abc hello world from cout! i tried: #include <io.h> #include <fcntl.h> _setmode(_fileno(stdout), _O_U8TEXT); problem: "cout" failed tried: std::locale mylocale(""); std::wcout.imbue

问题 This question already has answers here : Unexpected order of evaluation (compiler bug?) [duplicate] (3 answers) Closed 5 years ago . I'm new to C++ and study the increment and decrement operators. So I tried this example: int x = 4; cout << ++x << " " << x++ << " " << x++ << endl << endl; cout << x << endl; It returns this weird output on C++ .NET and QtCreator and 5 online C++ compilers: 7 5 4 7 The weird thing is that I expect something like this: 5 5 6 7 Can you explain what happens? 回答1:

问题 I understand PBYTE is unsigned char* from Windows Data Types. I want to be able to print all the contents, either using printf() or cout : PBYTE buffer; ULONG buffersize = NULL; serializeData(data, buffer, buffersize); //this function serializes "data" and stores the data in buffer and updates bufferSize).. Can you help me understand how to print this in C++? 回答1: If you want to print the memory address that buffer is pointing at, then you can do it like this: PBYTE buffer; ULONG buffersize =

问题 I understand PBYTE is unsigned char* from Windows Data Types. I want to be able to print all the contents, either using printf() or cout : PBYTE buffer; ULONG buffersize = NULL; serializeData(data, buffer, buffersize); //this function serializes "data" and stores the data in buffer and updates bufferSize).. Can you help me understand how to print this in C++? 回答1: If you want to print the memory address that buffer is pointing at, then you can do it like this: PBYTE buffer; ULONG buffersize =

问题 int a = 1, b = 2; int c = a*b + b==0; // c = 0 cout << a*b + b==0; // outputs 4 c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false. Why does putting the same expression in a cout statement output 4? 回答1: Because the precedence of these operators are operator* > operator+ > operator<< > operator== . Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0; . Then the

问题 int a = 1, b = 2; int c = a*b + b==0; // c = 0 cout << a*b + b==0; // outputs 4 c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false. Why does putting the same expression in a cout statement output 4? 回答1: Because the precedence of these operators are operator* > operator+ > operator<< > operator== . Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0; . Then the