Unscrambling C Declarations

This is a Java applet which can unscramble C declarations. The code to parse C declaration is translated from the C code described in K&R, 2nd edition, page 123.

Here are some C declarations from K&R, page 122. Try without the applet first and see if you can unscramble them.

Examples:

    int *f()
    int (*f)()
    char **argv
    int (*daytab)[13]

    void *comp()
    void (*comp)()
    char (*(*x())[])()
    char (*(*x[3])())[5]

Note: It can not handle declarations with function argument types or qualifiers like const, volatile, etc. For example, it can not handle declarations like:

    char * const (a[10]) (int **b)

Therefore, you have to rewrite the declaration to char *(a[10]) ()

Rules for unscrambling

Go to the leftmost identifier and say:
```
        identifier is
```
Look at left and right side of the identifier and apply the rules of precedence:
```
      ( )  => [ ] => *
```
Note: ( ) has the highest precedence and * has the lowest precedence. ( ) and [ ] are used as postfix operators and * is used as prefix operator.

if you see ( ), say:
```
    function returning
```
if you see [ ], say:
```
    array of
```
if you see *, say:
```
    pointer to
```

Example

Here is an example of the above rules. Let's unscramble it:

    char (*(*x[3])())[5]

Go to the leftmost identifier x and we say:
```
    x is
```
We look at the left and right side of x and see * is at the left side and [] is at the right side. As [] binds tighter than *, we say:
```
    array [0..2] of pointer to
```
Now, we see *x[3] is surrounded by () and * is on the left side and () is on the right side. As () has higher precedence than *, we say:
```
    function returning pointer to
```
Now, we look again and see, outside the parentheses, the char datatype is on the left side and [] is on the right side. So, we say:
```
    array [0..4] of char
```
And we're done. The result is:

x is array [0..2] of pointer to function returning pointer to array [0..4] of char

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