#include<iostream> #include<cstdlib> using namespace std; int main() { int temp = 300; cout << "Address of variable temp: " << (unsigned)&temp; return 0; } main.cpp: In function ‘int main()’: main.cpp:17:57: error: cast from ‘int*’ to ‘unsigned int’ loses precision...
I have a vector<int> and functions that only accepts a vector<unsigned int> references.I know that I could change/template the functions(and this is likely the best thing to do), but ideally I would have a way to cast/convert the vector<unsigned int> reference to a vector<int> referen...
int n2 = reinterpret_cast<int>(&o1); int n3 = reinterpret_cast<int&>(f1); int n4 = reinterpret_cast<int&>(o1); 2. 指针【引用】之间互转。如:float*转成int*、CBase&转成int&、CBase*转成CBase2*、CBase&转成CBase2&等 float f1 = 1.0f; CBase1 o1; int* n1 = reinterpret_cast<i...
应该是头文件没有包含。time函数加<ctime> srand和rand函数加<cstdlib> 你应该把整个文件贴出来。
unsigned int iTest = reinterpret_cast < unsigned int > (pIndepend); 也是没有任何问题的,运行效果一样,那我们能不能把指针转换为浮点型呢?这样: float fTest = reinterpret_cast < float > (pIndepend); double dTest = reinterpret_cast < double > (pIndepend); ...
error: cast from pointer to smaller type 'unsigned int' loses information 线程池代码上,往往需要输出对应的线程池编号,即(int) pthread_self() 但是在64为机器上可能导致因为int为4字节,指针统统为8字节,所以一般将int改成uintptr_t 即可解决。 即,将(unsigned int)强制类型转换修改成: uintptr_t 即可。
void func(unsigned char* in); I would like to pass a std::string to it, so I tried: Code: std::string temp = "Hello!"; func((unsigned char*)temp.c_str()); This works, but I thought it was good practice for type safety to use C++ style casting, yet reinterpret_cast does not...
4位的bunsignedintc:7;// 7位的c};intmain(){// 假设有一个整数,其中包含了按位打包的a、b...
hi, Please help me with the following code to get the difference in values. struct a{ int b1; int c1; char d1; } main() { unsigned int b=10; unsigned int c; c = b - (unsigned int )sizeof(a); printf("%d",c); } Here c returns some junk value. How can i get the......
显示:warning: int format, long int arg (arg 3)。警告原因: 象这样printf("%s%d, szDebugString, ulGwId);你的ulGwId是一个unsigned long型的,而你为它选择的输出形式却是 “%d”(这个格式是为整数型服务的-int)。解决方法: 这样的错误你只要做到参数类型一致就可以了,象上面的现象,你...