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I am trying to initalize a structure using braces, but i am really trying to initalize the structure that is pointed to by a pointer returned from a malloc call.
typedef struct foo{
int x;
int y;
} foo;
foo bar = {5,6};
I understand how to do that, but i need to do it in this context.
foo * bar = malloc(sizeof(foo));
*bar = {3,4};
269
int black = 1; int white = 5; int **board; int i; int j; int k; for(i=0;i<xsize;i++) { for(j=0;j<ysize;j++) { if(i<(xsize/2) && j<(ysize/2) && (i+j)%2!=
malloc allocates sizeof(struct node) bytes, and returns a void pointer to it, which we cast to struct node *. Under some conditions malloc could fail to allocate the required space, in which case it returns the special address NULL.
Create an Array of struct Using the malloc() Function in C The memory can be allocated using the malloc() function for an array of struct . This is called dynamic memory allocation. The malloc() (memory allocation) function is used to dynamically allocate a single block of memory with the specified size.
(This was answered in comments, so making it a CW).
You need to cast the right-hand side of the assignment, like so:
*bar = (foo) {3,4};
As pointed out by @cremno in the comment, this isn't a cast but rather an assignment of a compound literal
The relevant section of the C99 standard is: 6.5.2.5 Compound literals which says:
A postfix expression that consists of a parenthesized type name followed by a brace enclosed list of initializers is a compound literal. It provides an unnamed object whose value is given by the initializer list
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answered Sep 21 '22 11:09
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