如何使用Casablanca pplx::task
将参数传递给lambda函数?
没有卡萨布兰卡:
auto f = [](int x) {
cout <<"Hello "<<x<<endl;
return 5;
};
f(4);
与卡萨布兰卡一起,
auto g = [](/*argument x*/ ??) {
cout <<"Hello "<<x<<endl;
return 5;
};
pplx::task<int>(f(/*pass argument ?? */))
.then([](pplx::task<int> prevTask)
{
int res;
try {
res = prevTask.get();
}catch(const std::exception &e) {
cout << e.what() << endl;
}
}).wait();
我尝试传递简单的int
值pplx::task<int>(g(3))
,但无法编译。
代码:
#include <cpprest/asyncrt_utils.h>
using namespace std;
using namespace utility;
int main() {
auto f = [](int x) {
std::cout <<"Hello "<<x<<std::endl;
return 5;
};
f(4);
#if 0 // does not work
auto g = [](int x) {
cout <<"Hello "<<x<<endl;
return 5;
};
pplx::task<int>(g(3))
.then([](pplx::task<int> prevTask)
{
int res;
try {
res = prevTask.get();
}catch(const std::exception &e) {
cout << e.what() << endl;
}
}).wait();
#endif
}
CMakeLists.txt:
cmake_minimum_required (VERSION 3.5)
set(CMAKE_BUILD_TYPE Debug)
project(TEST)
message(STATUS "Compiling TestFile : ${PROJECT_NAME}")
set (CMAKE_CXX_STANDARD 17)
message(STATUS "Project Directory: ${PROJECT_SOURCE_DIR}")
set (CMAKE_BINARY_DIR ${CMAKE_SOURCE_DIR}/build)
set(EXECUTABLE_OUTPUT_PATH ${CMAKE_BINARY_DIR}/bin)
add_executable(${PROJECT_NAME} test.cpp)
set(REST_LIBRARIES "-lcpprest")
set(CMAKE_CXX_FLAGS "-std=c++17 -Wall -g -fsanitize=address")
target_link_libraries(${PROJECT_NAME} ${REST_LIBRARIES})
在Ubuntu libcpprest-dev
更新:它使用以下语法:
#include "cpprest/asyncrt_utils.h"
using namespace std;
using namespace utility;
int main() {
auto g = [](std::string x) {
cout<<x;
return std::string(" World!");
};
pplx::task_from_result<std::string>("Hello")
.then([g](std::string x)
{
auto ret = g(x);
return ret;
})
.then([](pplx::task<std::string> prevTask)
{
cout << prevTask.get() <<endl;
}).wait();
}
是否需要使用task_from_result
?