[英]Array Initialisation in C using Typedefs
Getting an odd error trying to initialise an array in C - anyone know why this might happen? 尝试在C中初始化数组时出现奇怪的错误 - 任何人都知道为什么会发生这种情况?
I have a global variable: 我有一个全局变量:
static my_type foo[6];
And in an included header file, I have: 在包含的头文件中,我有:
typedef uint32_t my_type[5];
I then in a function in the same file as the global variable try to do: 然后我在与全局变量相同的文件中的函数中尝试:
foo = {{1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, {1, 2, 3, 4, 8}, {1, 2, 3, 4, 9}, {1, 2, 3, 4, 10}};
The compiler (GCC4) gives the error 'expected expression before '{' token'. 编译器(GCC4)在'{'token'之前给出错误'expected expression'。
Anyone know what's gone wrong and how to fix it? 任何人都知道出了什么问题以及如何解决它?
Cheers! 干杯!
That's not initialization, that's assignment. 那不是初始化,那是分配。 Initialization has to be a single statement:
初始化必须是一个声明:
static my_type foo[6] = {{1, 2, 3, 4, 5},
{1, 2, 3, 4, 6},
{1, 2, 3, 4, 7},
{1, 2, 3, 4, 8},
{1, 2, 3, 4, 9},
{1, 2, 3, 4, 10}};
You cannot assign to a whole array in C89 with this syntax. 您无法使用此语法在C89中分配整个数组。 What you can do is
memcpy
from a const
: 你可以做的是来自
const
memcpy
:
void initialize_foo()
{
static const my_type init[6] =
{{1, 2, 3, 4, 5},
{1, 2, 3, 4, 6},
{1, 2, 3, 4, 7},
{1, 2, 3, 4, 8},
{1, 2, 3, 4, 9},
{1, 2, 3, 4, 10}};
assert(sizeof(foo) == sizeof(init));
memcpy(foo, init, sizeof(foo));
}
If you are under C99: 如果您在C99下:
ISO C99 supports compound literals.
ISO C99支持复合文字。 A compound literal looks like a cast containing an initializer.
复合文字看起来像包含初始值设定项的强制转换。 Its value is an object of the type specified in the cast, containing the elements specified in the initializer;
它的值是转换中指定类型的对象,包含初始值设定项中指定的元素; it is an lvalue.
这是一个左值。 As an extension, GCC supports compound literals in C89 mode and in C++.
作为扩展,GCC支持C89模式和C ++中的复合文字。
But foo
must be a pointer 但是
foo
必须是一个指针
#include <stdio.h>
#include <stdint.h>
typedef uint32_t my_type[5];
int main(void)
{
int i, j;
my_type *foo;
foo = ((my_type[]) {
{1, 2, 3, 4, 5},
{1, 2, 3, 4, 6},
{1, 2, 3, 4, 7},
{1, 2, 3, 4, 8},
{1, 2, 3, 4, 9},
{1, 2, 3, 4, 10}
});
for (i = 0; i < 6; i++) {
for (j = 0; j < 5; j++) {
printf("%d ", foo[i][j]);
}
printf("\n");
}
return 0;
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.