Using a pointer value in switch case.

Question

1.50/5 (2 votes)

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C++

#include <stdio.h>
int main() {
	//code
	int i=5;
	int *p=&i;
	printf ("%d", *p);
	switch (*p)
	{
	    case *p:
                printf ("&&");
                break;
	}
	return 0;
}

*p gives value of i. Then why is this program running into error?

What I have tried:

Switch statement works on integral expressions and pointer is not an integral expression. But i am passing the value pointed by the pointer that is 5. So should this not run?

Posted 21-Aug-18 10:04am

Member 13922884

Updated 2-Sep-18 3:02am

Nelek

v3

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3 solutions

Solution 2

Quote:
Using a pointer value in switch case.

Short answer: For what you want, it is if structure, not switch.

Posted 21-Aug-18 11:04am

Patrice T

Solution 1

From switch statement in C[^]

Quote:

The following rules apply to a switch statement

- The expression used in a switch statement must have an integral or enumerated type, or be of a class type in which the class has a single conversion function to an integral or enumerated type.

- You can have any number of case statements within a switch. Each case is followed by the value to be compared to and a colon.

- The constant-expression for a case must be the same data type as the variable in the switch, and it must be a constant or a literal.

- When the variable being switched on is equal to a case, the statements following that case will execute until a break statement is reached.

- When a break statement is reached, the switch terminates, and the flow of control jumps to the next line following the switch statement.

- Not every case needs to contain a break. If no break appears, the flow of control will fall through to subsequent cases until a break is reached.

- A switch statement can have an optional default case, which must appear at the end of the switch. The default case can be used for performing a task when none of the cases is true. No break is needed in the default case.

I think your pointer is not fitting in rule number one.

Posted 21-Aug-18 10:14am

Nelek

v3

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User 11060979 · Accepted Answer · 2018-09-02T03:02:00

Solution 3

This one switch (*p) is ok so far. But this one case *p: is the Problem.
Why?
Have a look to the Definition of the Switch Statement in c, e.g. here: switch Statement (C)[^]
The Definition says case constant-expression : and means you Need to use a constant Expression for the "case part", while *p is a (dereferenced pointer) variable and therefore not allowed here.

I hope this helps.

Posted 2-Sep-18 3:02am

User 11060979

Comments

Member 13922884 2-Sep-18 9:44am

Should *p be not replaced with its value it is referring to?
When executed does it not go as case 5: ... ?

[no name] 2-Sep-18 9:51am

Not sure that I understand correctly. But to make it work you should replace *p by the constant value 5:

case 5:

Again, the case value needs to be a constant value (in C) and therefore a variable is not allowed here.

Member 13922884 2-Sep-18 10:51am

Ohh.. my bad. I got your point. Thanks a lot.

[no name] 2-Sep-18 11:04am

You are very welcome, it helped also me to learn. Thank you for accepting.