#include void main() { unsigned char a,b; a=4|3; b=4&3; prin
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#include void main() { unsigned char a,b; a=4|3; b=4&3; printf("%d,%d",a,b); }
![#include void main() { unsigned char a,b; a=4|3; b=4&3; prin](/uploads/image/z/555180-60-0.jpg?t=%23include+void+main%28%29+%7B+unsigned+char+a%2Cb%3B+a%3D4%7C3%3B+b%3D4%263%3B+prin)
4和3的二进制
100 和
011
或的结果,有一位为1就为1
111
与的结果,都为1结果为1
100
011
---
000
这个结果为0
100 和
011
或的结果,有一位为1就为1
111
与的结果,都为1结果为1
100
011
---
000
这个结果为0
#include void main() { unsigned char a,b; a=4|3; b=4&3; prin
main() { unsigned char a,b; a=4|3; b=4&3; printf("%d %d\n",a
#include sbit a=P1^0; sbit b=P3^2; unsigned int n; void main
有以下程序 main( ) { unsigned char a,b; a=4|3; b=4&3; printf(“%d
#include main() { unsigned char a=8,c; c=a>>3; printf("%d\n"
void fun (char *a,char *b) {a=b} void main() {char c1='A',c2
#include void main( ) { char c1='a',c2='b',c3='c',c4='\101',
#include void main() {char c1='a',c2='b',c3='c',c4='
#include void main( ) { int a=3,b=5; int *p,*q; void f1(int
C语言菜鸟问题:如下#include "stdio.h"void main(){ char a='A'+'8'-'4';
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char point(char*p) {p+=3;return *P} main() {char b[4]={'a','