问题描述
我正在尝试声明一个指针并将该指针传递给分配了内存的函数.这是一个最小的例子:
I'm trying to declare a pointer and pass that pointer to a function where memory is allocated. Here is a minimal example:
#include <string>
#include <iostream>
using namespace std;
void alloc_mem(int &size, double *x);
int main()
{
double *X;
int imax;
alloc_mem(imax, X);
cout << "imax = " << imax << endl;
for (int i = 0; i < imax; i++) {
cout << "X = " << X[i] << endl;
}
delete[]X;
return 0;
}
void alloc_mem(int &size, double *x)
{
size = 10;
x = new double[size];
for (int i = 0; i < size; i++) {
x[i] = (double)i;
}
}
这段代码可以编译,但是当我尝试打印 X 的值时出现分段错误.我知道我没有正确地将变量传递到函数中,但我不知道该怎么做.我相信我正在对 x 的副本进行操作.
This code compiles, but I get a segmentation fault when I try to print out the values of X. I know that I'm not passing the variable into the function correctly, but I'm not sure how to do it. I believe that I'm operating on a copy of x.
此外,编写此代码是为了重现我在更大的代码中遇到的问题.
Also, this code was written to reproduce an issue I'm having in a much larger code.
推荐答案
参数 double *x
是函数 alloc_mem 的局部变量.当函数结束执行时,变量将被销毁.main 中的原始变量 X 不知道对这个参数做了什么,因为它是按值传递的,它是函数中使用的它的副本.
Parameter double *x
is a local variable of function alloc_mem. When the function will end its execution the variable will be destroyed. The original variable X in main knows nothing what was done with this parameter because it was passed by value that is a copy of it was used in the function.
要么通过指针传递指针,要么通过引用传递指针.例如
Either pass the pointer by pointer or by reference. For example
void alloc_mem(int &size, double **x);
void alloc_mem(int &size, double * &x);
void alloc_mem(int &size, double **x)
{
size = 10;
*x = new double [size];
for ( int i = 0; i < size; i++ ) ( *x )[i] = i;
}
void alloc_mem(int &size, double * &x)
{
size = 10;
x = new double [size];
for ( int i = 0; i < size; i++ ) x[i] = i;
}
对于我来说,我会用以下方式定义函数
As for me I would define the function the following way
double * alloc_mem( int &size )
{
size = 10;
x = new double [size];
for ( int i = 0; i < size; i++ ) x[i] = i;
return x;
}
如果在调用函数之前就知道大小,那么它可以写得更简单
if size is known before calling the function then it could be written even simpler
double * alloc_mem( int size )
{
x = new double [size];
for ( int i = 0; i < size; i++ ) x[i] = i;
return x;
}
考虑那个循环
for ( int i = 0; i < size; i++ ) x[i] = i;
可以代替标准算法std::iota
例如
can be substituted for standard algorithm std::iota
For example
std::iota( x, x + size, 0.0 );