我在C中有一个大数组(不是c++,如果这有区别的话)。我想初始化相同值的所有成员。
我发誓我曾经知道一个简单的方法。在我的情况下,我可以使用memset(),但没有一种方法来做到这一点,这是建立在C语法?
我在C中有一个大数组(不是c++,如果这有区别的话)。我想初始化相同值的所有成员。
我发誓我曾经知道一个简单的方法。在我的情况下,我可以使用memset(),但没有一种方法来做到这一点,这是建立在C语法?
当前回答
如果你的编译器是GCC,你可以使用以下“GNU扩展”语法:
int array[1024] = {[0 ... 1023] = 5};
查看详细描述: http://gcc.gnu.org/onlinedocs/gcc-4.1.2/gcc/Designated-Inits.html
其他回答
我看到了一些使用这种语法的代码:
char* array[] =
{
[0] = "Hello",
[1] = "World"
};
它变得特别有用的地方是当你创建一个使用枚举作为索引的数组时:
enum
{
ERR_OK,
ERR_FAIL,
ERR_MEMORY
};
#define _ITEM(x) [x] = #x
char* array[] =
{
_ITEM(ERR_OK),
_ITEM(ERR_FAIL),
_ITEM(ERR_MEMORY)
};
这使事情保持有序,即使您碰巧把一些枚举值写成了无序的顺序。
关于这项技术的更多信息可以在这里和这里找到。
没有人提到访问初始化数组元素的索引顺序。我的示例代码将为它提供一个说明性示例。
#include <iostream>
void PrintArray(int a[3][3])
{
std::cout << "a11 = " << a[0][0] << "\t\t" << "a12 = " << a[0][1] << "\t\t" << "a13 = " << a[0][2] << std::endl;
std::cout << "a21 = " << a[1][0] << "\t\t" << "a22 = " << a[1][1] << "\t\t" << "a23 = " << a[1][2] << std::endl;
std::cout << "a31 = " << a[2][0] << "\t\t" << "a32 = " << a[2][1] << "\t\t" << "a33 = " << a[2][2] << std::endl;
std::cout << std::endl;
}
int wmain(int argc, wchar_t * argv[])
{
int a1[3][3] = { 11, 12, 13, // The most
21, 22, 23, // basic
31, 32, 33 }; // format.
int a2[][3] = { 11, 12, 13, // The first (outer) dimension
21, 22, 23, // may be omitted. The compiler
31, 32, 33 }; // will automatically deduce it.
int a3[3][3] = { {11, 12, 13}, // The elements of each
{21, 22, 23}, // second (inner) dimension
{31, 32, 33} }; // can be grouped together.
int a4[][3] = { {11, 12, 13}, // Again, the first dimension
{21, 22, 23}, // can be omitted when the
{31, 32, 33} }; // inner elements are grouped.
PrintArray(a1);
PrintArray(a2);
PrintArray(a3);
PrintArray(a4);
// This part shows in which order the elements are stored in the memory.
int * b = (int *) a1; // The output is the same for the all four arrays.
for (int i=0; i<9; i++)
{
std::cout << b[i] << '\t';
}
return 0;
}
输出结果为:
a11 = 11 a12 = 12 a13 = 13
a21 = 21 a22 = 22 a23 = 23
a31 = 31 a32 = 32 a33 = 33
a11 = 11 a12 = 12 a13 = 13
a21 = 21 a22 = 22 a23 = 23
a31 = 31 a32 = 32 a33 = 33
a11 = 11 a12 = 12 a13 = 13
a21 = 21 a22 = 22 a23 = 23
a31 = 31 a32 = 32 a33 = 33
a11 = 11 a12 = 12 a13 = 13
a21 = 21 a22 = 22 a23 = 23
a31 = 31 a32 = 32 a33 = 33
11 12 13 21 22 23 31 32 33
对于延迟初始化(即类成员构造函数初始化),请考虑:
int a[4];
unsigned int size = sizeof(a) / sizeof(a[0]);
for (unsigned int i = 0; i < size; i++)
a[i] = 0;
我知道最初的问题明确地提到了C而不是c++,但如果你(像我一样)来这里寻找c++数组的解决方案,这里有一个巧妙的技巧:
如果你的编译器支持fold表达式,你可以使用template magic和std::index_sequence生成一个包含你想要的值的初始化列表。你甚至可以把它想象成一个老板:
#include <array>
/// [3]
/// This functions's only purpose is to ignore the index given as the second
/// template argument and to always produce the value passed in.
template<class T, size_t /*ignored*/>
constexpr T identity_func(const T& value) {
return value;
}
/// [2]
/// At this point, we have a list of indices that we can unfold
/// into an initializer list using the `identity_func` above.
template<class T, size_t... Indices>
constexpr std::array<T, sizeof...(Indices)>
make_array_of_impl(const T& value, std::index_sequence<Indices...>) {
return {identity_func<T, Indices>(value)...};
}
/// [1]
/// This is the user-facing function.
/// The template arguments are swapped compared to the order used
/// for std::array, this way we can let the compiler infer the type
/// from the given value but still define it explicitly if we want to.
template<size_t Size, class T>
constexpr std::array<T, Size>
make_array_of(const T& value) {
using Indices = std::make_index_sequence<Size>;
return make_array_of_impl(value, Indices{});
}
// std::array<int, 4>{42, 42, 42, 42}
constexpr auto test_array = make_array_of<4/*, int*/>(42);
static_assert(test_array[0] == 42);
static_assert(test_array[1] == 42);
static_assert(test_array[2] == 42);
static_assert(test_array[3] == 42);
// static_assert(test_array[4] == 42); out of bounds
您可以查看工作中的代码(在Wandbox中)
如果数组声明为静态或全局,则所有元素 在数组中已经有默认默认值0。 一些编译器在调试模式中将array的默认值设置为0。 将默认值设置为0很容易: Int数组[10]= {0}; 但是,对于其他值,您必须使用memset()或loop;
例子: int数组[10]; Memset(数组,- 1,10 *sizeof(int));