[英]Initializing an structure pointer with -> operator
我必須使用指向它的點來初始化結構,以這種方式,upper_left是點(10,25)和lower_right(20,15)
struct point {int x, y;};
struct rectangle {struct point upper_left, lower_right;};
struct rectangle *p;
p = malloc(sizeof(*p));
p->upper_left.x = 10;
p->upper_left.y = 25;
p->lower_right.x = 20;
p->lower_right.y = 15;
有什么方法可以使它更“緊湊”,而不是一一對應嗎? 我已經嘗試過了,但是編譯器的錯誤是相同的“'{'令牌之前的預期表達式”
p->upper_left = {10, 25};
p->lower_right = {20,15};
/////////////
p->upper_left = {.x = 10, .y = 25};
p->lower_right = {.x = 20, .y = 15};
//////////////////////////////
*p = {.upper_left = {10, 25}, .lower_right = {20, 15}};
是復合文字
p->upper_left = (struct point){10, 25};
*p = (struct rectangle){.upper_left = {10, 25}, .lower_right = {20, 15}};
您可以在這里進行實驗: https : //godbolt.org/z/yw_JC0
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