int x=20;printf(“%d”,0
int x=20;printf(“%d”,0
int x=20; printf("%d\n",x);
一个简单的程序题#includeint main(void){int x=20;printf("%d",0
int x=10,y=20; main() {func();printf("%d,%d\n",x,y);} func()
int a=10;f1(){int a=20;printf("%d",a);}f2(){printf("%d",a);}
int f(int x){if(x==0) return 1;z+x*f(x-1);printf("%d",z);ret
void main() {int x=4;if(x++>=5)printf("%d/n",x);else printf(
main() { int x=0; sub(&x,8,1); printf("%d\n",x); } sub(int *
main() { int x=10; int y=x++; printf("%d,%d\n",(x++,y),y++);
main() { int x=2002,y=2003; printf("%d\n",(x,y)); } 为什么结果是20
int i=9999;printf("%d\n",printf("%d",printf("%d",printf("%d"
void main() { int x=0,y=5,z=3; while() printf("%d,%d,%d\n",x