求下面的程序的结果~void main( ){Stack S;Char x,y;InitStack(S);X=’c’;y=’k’;Push(S,x); Push(S,’a’); Push(S,y);Pop(S,x); Push(S,’t’); Push(S,x);Pop(S,x); Push(S,’s’);while(!StackEmpty(S)){ Pop(S,y);printf(y); };Printf(x);}

求下面的程序的结果~void main( ){Stack S;Char x,y;InitStack(S);X=’c’;y=’k’;Push(S,x); Push(S,’a’); Push(S,y);Pop(S,x); Push(S,’t’); Push(S,x);Pop(S,x); Push(S,’s’);while(!StackEmpty(S)){ Pop(S,y);printf(y); };Printf(x);}
求下面的程序的结果~
void main( ){
Stack S;
Char x,y;
InitStack(S);
X=’c’;y=’k’;
Push(S,x); Push(S,’a’); Push(S,y);
Pop(S,x); Push(S,’t’); Push(S,x);
Pop(S,x); Push(S,’s’);
while(!StackEmpty(S)){ Pop(S,y);printf(y); };
Printf(x);
}

求下面的程序的结果~void main( ){Stack S;Char x,y;InitStack(S);X=’c’;y=’k’;Push(S,x); Push(S,’a’); Push(S,y);Pop(S,x); Push(S,’t’); Push(S,x);Pop(S,x); Push(S,’s’);while(!StackEmpty(S)){ Pop(S,y);printf(y); };Printf(x);}
根据堆栈的“后进先出”的特性,我们可以得到如下过程：
void main( ){
Stack S;
Char x,y;
InitStack(S);
x=’c’;y=’k’; //x='c',y='k'
Push(S,x); //x='c',y='k',S="c"
Push(S,’a’); //x='c',y='k',S="ac"
Push(S,y); //x='c',y='k',S="kac"
Pop(S,x); //x='k',y='k',S="ac"
Push(S,’t’); //x='k',y='k',S="tac"
Push(S,x); //x='k',y='k',S="ktac"
Pop(S,x); //x='k',y='k' S="tac"
Push(S,’s’); //x='k',y='k' S="stac"
while(!StackEmpty(S))
{
Pop(S,y); //依次为y='s',y='t',y='a',y='c'
printf(y); //打印依次为s,t,a,c
}
Printf(x);//x='k'
}