CODEWITHSUNDEEP
cswitch-statement
Liu
Liu

Reputation: 433

variable defined in switch in C

When I tried to assign a pointer in switch statement, the
compile gave me an error warning, but when I debugged the code by declaring
the pointer outside the switch statement. The code
syntactically passed. Why? Can anyone explain it to me? Thank you!

Before I debugged:

void Left_Balance(treeNode **T)
{
    treeNode *p=(*T)->left;  
    switch(p->bf)
    {
        case LH:
            p->bf=(*T)->bf=EQ;
            L_Rotate(T);
            break;
        case RH:
             treeNode *q = p->right;  
            switch(q->bf)
           {
               case LH:
                   (*T)->bf=RH;
                   p->bf=EQ;
                   break;
               case EQ:
                   (*T)->bf=q->bf=EQ;
                   break;
               case RH:
                   p->bf=LH;
                   (*T)=EQ;
                   break;
           }
            q->bf=EQ;
            R_Rotate(&p);
            L_Rotate(T);
            break;
    }
}

After I debugged :

void Left_Balance(treeNode **T)
{
    treeNode *p=(*T)->left;
    treeNode *q;     //Here is the difference 
    switch(p->bf)
    {
        case LH:
            p->bf=(*T)->bf=EQ;
            L_Rotate(T);
            break;
        case RH:
            q = p->right;    //Here is the difference 
            switch(q->bf)
           {
               case LH:
                   (*T)->bf=RH;
                   p->bf=EQ;
                   break;
               case EQ:
                   (*T)->bf=q->bf=EQ;
                   break;
               case RH:
                   p->bf=LH;
                   (*T)=EQ;
                   break;
           }
            q->bf=EQ;
            R_Rotate(&p);
            L_Rotate(T);
            break;
    }
}

Upvotes: 0

Views: 67

Answers (2)

dbush
dbush

Reputation: 225344

    case RH:
         treeNode *q = p->right;

Here you have a label right before a declaration. A label can only come before a statement. You can fix this by adding an empty statement right before the declaration that the label can be associated with:

    case RH:
         ;
         treeNode *q = p->right;

Upvotes: 0

bruno
bruno

Reputation: 32596

in your first version treeNode *q = p->right; is not in a block ({})

an other possibility was to add that block

    case RH:
  {
    treeNode *q = p->right;  
        switch(q->bf)
       {
           case LH:
               (*T)->bf=RH;
               p->bf=EQ;
               break;
           case EQ:
               (*T)->bf=q->bf=EQ;
               break;
           case RH:
               p->bf=LH;
               (*T)=EQ;
               break;
       }
        q->bf=EQ;
        R_Rotate(&p);
        L_Rotate(T);
        break;
  }

Warning 

 (*T)=EQ;

is very suspicious because *T is a treeNode * and EQ is not one because of case EQ

Probably you want

 (*T)->bf = EQ;

(edit)

As @dbush indicates it is enough to add a statement between the label and the declaration, but to not put the declaration in a block is very dangerous, look at that code :

#include <stdio.h>

int main(int argc, char ** argv)
{
  switch (argc) {
  case 1:
    puts("case 1");

    int i = 123;

    puts("i initialized and continue in case 2");

  case 2:
    printf("in case2, i values : %d\n", i);
  }

  return 0;
}

What about i used in printf if the program is called with an argument initializing argc is 2 ?

Note gcc accepts that code but not g++ without the option -fpermissive and with that option the code is compiled without case 2:

This is why I proposed to add a block.

Upvotes: 1

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