Nettet21. jun. 2010 · int a=10,b=4,c=20,d=6;System.out.println(a++*b+c*--d) 我觉得是10*4+20*5=140， 可答案是164，为什么呢 Nettet12. okt. 2024 · So the // value of expression a-- is 1. Since the // first operand of logical and is 1, // shortcircuiting doesn't happen here. So // the expression --b is executed and --b …

void main() { int a=10, b; b = a++ + ++a; printf("%d %d %d %d", b, a++ ...

Nettet25. nov. 2024 · So, ‘a++’ will be executed first as it is the last parameter of the first printf () statement. It will print 10. Although, now the value has been increased by 1, so the second last argument, i.e., will print 11. Similarly, the other statements will also get executed. Note: In pre-increment, i.e., ++a, it will increase the value by 1 before ... Nettet请写出下列代码的输出内容。. __牛客网. ++i表示，i自增1后再参与其它运算；而i++ 则是i参与运算后，i的值再自增1。. 有一点疑惑请教一下：b=a++这句不考虑运算符的优先级了么？. 自增运算符的优先级不是比赋值的优先级高吗？. 后置版本（i++）：也会将运算 ... the art story rufino tamayo

Operators in C - GeeksQuiz - GeeksForGeeks

Nettet#include int main() { int a=4,b,c; b = --a; c = a--; printf("%d %d %d",a,b,c); return 0; } a) 3 3 2 b) 2 3 2 c) 3 2 2 d) 2 3 3 View Answer Answer:- d) 2 3 3 The first expression is b=–a; so, a becomes 3 (a=3) and b=3. Nettetint a=1; // initialization int b=0; // initialization b=++a + ++a; // find the pre-increment i.e. 2 increments of 'a' so now 'a' in this step will be incremented by 2 so now 'a' will contain 1+2=3. so now a=3. Again before assignment compute 'a+a' which is '3+3'=6 printf("%d %d",a,b); //3 6} Just a trick:- always compute the pre-increments in ... NettetIncrement and decrement operators are unary operators that add or subtract one, to or from their operand, sequentially. They are commonly implemented in imperative … the glengariff sea point