以下程序的输出结果是 #include void prt (int *x,int*y,int*z){ printf("%d,%d,%d\n",++*x,++*y,*(z++));}main() { int a=10,b=40,c=20; prt (&a,&b,&c); prt (&a,&b,&c); }A)11,42,31 12,22,41 B)11,41,20 12,42,20 C)11,2
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![以下程序的输出结果是 #include void prt (int *x,int*y,int*z){ printf(](/uploads/image/z/6033955-67-5.jpg?t=%E4%BB%A5%E4%B8%8B%E7%A8%8B%E5%BA%8F%E7%9A%84%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C%E6%98%AF+%23include+void+prt+%EF%BC%88int+%2Ax%2Cint%2Ay%2Cint%2Az%EF%BC%89%7B+printf%EF%BC%88%22%25d%2C%25d%2C%25d%EF%BC%BCn%22%2C%2B%2B%2Ax%2C%2B%2B%2Ay%2C%2A%EF%BC%88z%2B%2B%EF%BC%89%EF%BC%89%3B%7Dmain%EF%BC%88%EF%BC%89+%7B+int+a%3D10%2Cb%3D40%2Cc%3D20%3B+prt+%EF%BC%88%26a%2C%26b%2C%26c%EF%BC%89%3B+prt+%EF%BC%88%26a%2C%26b%2C%26c%EF%BC%89%3B+%7DA%EF%BC%8911%2C42%2C31+12%2C22%2C41+B%EF%BC%8911%2C41%2C20+12%2C42%2C20+C%EF%BC%8911%2C2)
以下程序的输出结果是 #include void prt (int *x,int*y,int*z){ printf("%d,%d,%d\n",++*x,++*y,*(z++));}main() { int a=10,b=40,c=20; prt (&a,&b,&c); prt (&a,&b,&c); }A)11,42,31 12,22,41 B)11,41,20 12,42,20 C)11,2
以下程序的输出结果是
#include void prt (int *x,int*y,int*z)
{ printf("%d,%d,%d\n",++*x,++*y,*(z++));}
main() { int a=10,b=40,c=20; prt (&a,&b,&c); prt (&a,&b,&c); }
A)11,42,31 12,22,41 B)11,41,20 12,42,20
C)11,21,40 11,21,21 D)11,41,21 12,42,22
以下程序的输出结果是 #include void prt (int *x,int*y,int*z){ printf("%d,%d,%d\n",++*x,++*y,*(z++));}main() { int a=10,b=40,c=20; prt (&a,&b,&c); prt (&a,&b,&c); }A)11,42,31 12,22,41 B)11,41,20 12,42,20 C)11,2
void prt (int *x,int*y,int*z)
{
\x09printf("%d,%d,%d\n",++*x,++*y,*(z++));
// x和y用" * "符号取形参地址的值.进行前置自加,而*(z++),分解为(z++)是将z的值先用 * 符号取出来输出,在将地址加1,所以z的值不变,而x和y进行自加运算
}
int main() {
\x09 int a=10,b=40,c=20;
\x09 prt (&a,&b,&c); //传递给函数变量的地址
\x09 prt (&a,&b,&c);
return 0;
}
答案为:B