求下面的程序的结果~void main( ){Stack S;Char x,y;InitStack(S);X=’c’;y
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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);
}
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](/uploads/image/z/8759552-32-2.jpg?t=%E6%B1%82%E4%B8%8B%E9%9D%A2%E7%9A%84%E7%A8%8B%E5%BA%8F%E7%9A%84%E7%BB%93%E6%9E%9C%7Evoid+main%28+%29%7BStack+S%3BChar+x%2Cy%3BInitStack%28S%29%3BX%3D%E2%80%99c%E2%80%99%3By)
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'
}
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'
}
求下面的程序的结果~void main( ){Stack S;Char x,y;InitStack(S);X=’c’;y
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