I was wondering if it is possible in C to initialise a structure in a following way:
struct Test* test = { .int_value = 10, .char_value = 'c', .pointer_to_float_value = (float*)1.512 };
If I try to do this with a structure defined in a way:
struct Test
{
int int_value;
char char_value;
float* pointer_to_float_value;
};
I get an error for all elements of the structure:
error: field name not in record or union initializer
Is there a way to bypass this issue?
It is indeed possible, but you are declaring a pointer to a struct and trying to initialize it as a struct (not pointer). The same issue with the pointer to float. The following should work:
float a = 1.512;
struct Test test_struct = { .int_value = 10, .char_value = 'c', .pointer_to_float_value = &a };
struct Test *test = &test_struct;
This syntax
struct Test* test { .int_value = 10, .char_value = 'c', .pointer_to_float_value = (float*)1.512 };
is invalid.
Instead you could use a compound literal as for example
float f = 1.512f;
struct Test* test = &( struct Test ){ .int_value = 10, .char_value = 'c', .pointer_to_float_value = &f };
From the C Standard (6.5.2.5 Compound literals)
5 The value of the compound literal is that of an unnamed object initialized by the initializer list. If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.
There are a number of problems with your code. You don't have a = sign and you are not handling pointers and types correctly.
It can be done like:
struct Test * test = &(struct Test){ .int_value = 10,
.char_value = 'c',
.pointer_to_float_value = &(float){1.512f} };
Notice how it uses (struct Test) to set the type of the compound literal and & to get a pointer to it. The same applies to the float pointer.