int x=3,y=4; printf("%d,%d",(x.y),(y,x));

int x=3,y=4; printf(%d,%d,(x.y),(y,x)); void fun(int *x,int y){printf(%d%d,*x,*y);*x=3;*y=4;}main(){int x=1,y=2;fun(&y,&x);printf(%d%d,x,y);} #include main() { int x=3,y; do { y=x--; if(!y) {printf(*);continue;} printf(#); }whi下面程序段____________#include main(){int x=3,y;do{y=x--;if(!y){printf(*);continue;}printf(#);}while(1 void fun(int x,int y) {x=x+y;y=x-y;x=x-y; printf(%d,%d,,x,y); } main() {int x=2,y=3; fun(x,y); privoid fun(int x,int y){x=x+y;y=x-y;x=x-y;printf(%d,%d,,x,y);}main(){int x=2,y=3;fun(x,y);printf(%d,%d ,x,y);} # include void p(int *x,int y){ ++*x; y=y+2; } void main() { int x=2,y=3; p(&y,y); printf(# includevoid p(int *x,int y){++*x;y=y+2;}void main(){int x=2,y=3;p(&y,y);printf(%d#%d,x,y);} int func(int x,int y ) { return(x+y) } main() {int a=1,b=2,c=3,d=4,e=5;printf(&d ,func((a+b,b+c,c+a),(d+e))); #include void main( ) { int a=3,b=5; int *p,*q; void f1(int x,int y);void f2(int *x,int *y);p=&a; q=&b;f1(*p,*q) ;printf(a=%d ,b=%d ,a,b);f2(p,q);printf(a=%d ,b=%d ,a,b);}void f1(int x,int y) { int t;t=x; x=y; y=t;}void f2(int *x,int *y) { int 若有说明int x=10,y=20;请写出各printf语句的输出结果 printf(“%3o ”,x%y,x,y) #include int main() { int max; int x,y,z; if x>y printf(max=x); else printf(max);if max C语言中怎么计算x,y的值?#define int main(void){int x,y;x=y+1;y=3+25;printf( )} 程序有个部分读不懂,#include void main(){int x,y,t;double a;float b;int c;scanf(%d %d,&x,&y);c=b=a=20/3;t=(x%y,x/y);printf(%d %d ,x--,--y); printf(%d ,t); printf(%d ,(x=5*6,x*4,x+5));printf(%f %f %f ,c,b,a);}中的t=(x%y,x/y)和 C语言 答案为什么是8 4 #includevoid f(int y,int *x){y=y+*x;*x=*x+y;}main(){int x=2,y=4;f(y,&x);printf(%d %d ,x,y);} #includevoid f(int y,int *x){ y=y+*x; *x=*x+y;} void main(){ int x=2,y=4;f(y,&x);printf(%d,%d ,x,y);} int x,y; x=13; y=5; {printf(%d,x%=(y/=2)); void f( int y,int *x) {y=y+*x; *x=*x+y;} main( ) { int x=2,y=4; f(y,&x); printf(%d %d ,x,y); } void f( int y,int *x){y=y+*x; *x=*x+y;}main( ){ int x=2,y=4;f(y,&x);printf(%d %d ,x,y);} 执行后输出的结果是 . #include struct ord { int x,y;} dt[2]={1,2,3,4}; main() {struct ord *p=dt; printf(%d,,++#includestruct ord {int x,y;} dt[2]={1,2,3,4};main(){struct ord *p=dt;printf(%d,,++p->x);printf(%d,,++p->y);}++p->x++p->y int x=20;printf(“%d”,0 int x=20;printf(“%d”,0