int x=10,y=20; main() { int k,y=5; k=f1(); printf("k=%d,x=%d,y=%d",k,x,y); k=f2();

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/31 01:46:25
intx=10,y=20;main(){intk,y=5;k=f1();printf("k=%d,x=%d,y=%d",k,x,y);k=f2();intx=10,y=20;main(){intk,y

int x=10,y=20; main() { int k,y=5; k=f1(); printf("k=%d,x=%d,y=%d",k,x,y); k=f2();
int x=10,y=20; main() { int k,y=5; k=f1(); printf("k=%d,x=%d,y=%d",k,x,y); k=f2();

int x=10,y=20; main() { int k,y=5; k=f1(); printf("k=%d,x=%d,y=%d",k,x,y); k=f2();
大概意思是这样的:f1()与f2()是个函数,在执行printf前,把f1()函数返回的值赋给k.在执行完printf后再把函数f2()的值赋给k.

823456

你想问什么?

下列程序运行结果为() main() { int x=20,y; y=2 int x=10,y=20; main() {func();printf(%d,%d ,x,y);} func() {int z; z=x;x=y;y=z;x=2*x;y=2+y;}的运算结果是 #include int main(){ int x=0,y=0; while(x #include void main() { int a=3,b=2,c=1; int x=10,y=20; if(a main(){ int x=5; while(x main() { int x=10; int y=x++; printf(%d,%d ,(x++,y),y++); }请先说出你的答案! #include #include int main() { int x,y,sqrt; scanf (%d,&x); y=sqrt(x); if (x> int x=10,y=20; main() { int k,y=5; k=f1(); printf(k=%d,x=%d,y=%d,k,x,y); k=f2(); public class Simple{public static void main(String args[]){simple(10);simple(int x){int y=x*x;System.out.println(y)}}}请帮我改正一下 main( ) { int x=010,y=10,z=0x10; printf(“%d,%d,%d ”,x,y,z); } main( ) { int x=010,y=10,z=0x10; printf(“%d,%d,%d ”,x,y,z); } #include using namespace std; int main() { int a,b,c; a=3; int f(int x,int y,int z);#include using namespace std; int main() {int a,b,c;a=3;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout # 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);} main() { int x=2002,y=2003; printf(%d ,(x,y)); } 为什么结果是2003? main() { int x=2002,y=2003; printf(%d ,(x,y)); } 求高手解释一下这段程序#include class Test {int x,y; public: Test(int i,int j=0) {x=i;y=j;} int get(int i,int j) {return i+j;} }; void main() {Test t1(2),t2(4,6); int (Test::*p)(int,int=10); p=Test::get; cout struct st{int x,int*y;}*p; int s[]={5,6,7,8} st a[]={10,&s[0],20,&s[1]30,&[2],40,&s[3]} main( ) {p=a;cout main(){int a[]={2,4,6,8,10};int y=1,x,*p;p=&a[1];for(x=0;x