[英]How can I statically initialise an array of structs containing a union?
I'm porting some old code from C to C++ in Visual Studio 2010 and I came across this: 我正在Visual Studio 2010中将一些旧代码从C移植到C ++,并且遇到了这个问题:
typedef struct OptionDef {
const char *name;
int flags;
union {
void *dst_ptr;
int (*func_arg)(void *, const char *, const char *);
size_t off;
} u;
const char *help;
const char *argname;
} OptionDef;
static const OptionDef options[] = {
{ "x", HAS_ARG, { .func_arg = opt_width }, "force displayed width", "width" },
...
Which now fails with a syntax error. 现在由于语法错误而失败。 I've seen the response for Statically initialize anonymous union in C++ but overloading the constructors won't work because I'm setting up an array. 我已经看到了C ++中对静态初始化匿名联合的响应,但是重载构造函数将不起作用,因为我正在设置数组。 Is there any other way of doing this (rather than just rewriting the code not to use a union)? 还有其他方法(而不是仅重写代码以不使用联合)吗?
Update: I should have been more specific - the array contains different initialisers using all parts of the union: 更新:我应该更具体一些-数组包含使用联合所有部分的不同初始化程序:
static int is_full_screen;
{ "fs", OPT_BOOL, { &is_full_screen }, "force full screen" },
So just changing the order of the union won't help. 因此,仅更改工会的顺序将无济于事。
C++ does not have the .member
initialization syntax that C has. C ++没有C具有的.member
初始化语法。
You can use aggregate initialization with unions but only on the first member. 您可以对联合使用聚合初始化,但只能在第一个成员上使用。
Thus, rewrite it with the one you want to set as the first member: 因此,使用您要设置为第一个成员的成员重写它:
union {
int (*func_arg)(void *, const char *, const char *);
void *dst_ptr;
size_t off;
} u;
static const OptionDef options[] = {
{ "x", HAS_ARG, { opt_width }, "force displayed width", "width" },
You could also give your struct a constructor - C++11 should allow you to use brace initializers. 您还可以为您的结构提供构造函数-C ++ 11应该允许您使用括号初始化程序。
Example: 例:
struct foo {
int flags;
struct uwrap {
uwrap(int (*func_arg)(void *, const char *, const char *))
: func_arg(func_arg) {}
uwrap(int off)
: off(off) {}
union {
void *dst_ptr;
int (*func_arg)(void *, const char *, const char *);
int off;
};
} u;
};
int func(void *, const char *, const char *) {}
int main() {
foo f[] = { { 1, {func}}, { 2, {0}} };
}
In C++03 you can do it with temporaries if the struct has a constructor: 在C ++ 03中,如果该结构具有构造函数,则可以使用临时函数:
foo f[] = { foo(1, func), foo(3, 0) };
Just do this: 只要这样做:
static const OptionDef options[] = {
{ "x", HAS_ARG, {opt_width }, "force displayed width", "width" },
...
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