1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 | #include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <assert.h>
typedef struct {
int x;
int y;
} Point;
void print_point(const char* name, Point p) {
printf("%s is a point at x=%d, y=%d\n", name, p.x, p.y);
}
int main(int argc, char* argv[]) {
// NAMED INITIALIZER
Point p1 = { .x = 5, .y = 6 };
//└───────┬────────┘
// named initializer
Point p2 = p1; // regular initializer
printf("p2 is a point at x=%d, y=%d\n", p2.x, p2.y);
// Suppose we want to set p1 to x=3, y=4.
// WRONG WAY - named initializer is only for initializing (initial value) on the line
// where the variable is declared.
// p1 = { .x = 3, .y = 4 }; // This is not a declaration. It is an assignment.
// RIGHT WAY: COMPOUND LITERAL
p1 = (Point) { .x = 3, .y = 4 };
print_point("p1", p1);
print_point("Mr. Compound Literal", (Point) { .x = 500, .y = 600 });
// For an assignment, this is equivalent to assigning each field individually.
// p1.x = 3
// p1.y = 4
// Easier to notice if you missed a field.
// You could have a compound literal nested inside of another.
// You could pass a compound literal to a function.
return EXIT_SUCCESS;
}
/*
compound_literal.c: In function ‘main’:
compound_literal.c:25:7: error: expected expression before ‘{’ token
p1 = { .x = 3, .y = 4 }; // This is not a declaration. It is an assignment.
^
*/
/* vim:set tabstop=4 shiftwidth=4 fileencoding=utf-8 noexpandtab: */
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