Why array is not initialized inside if-statement?

Question

Why array is not initialized inside if-statement?

Debojit Acharjee 455

I want to know why the text is not initialized to the array in the program below.

#include <stdio.h>

char a[] = "RED";

int main ()
{

int b=-1, c=0;

printf("Enter number: ");
//scanf("%d", b);

if (b < 0)
{

char a[] = "GREEN";
c = 1;

}

char a[] = "BLUE";

    printf("%d, %s", c, a);

    return 0;
}

In this program the if stamen is fired by the variable 'b', but why the array 'a' is not assigned the text "GREEN"? The output shows only the value of c.

Barry Schwarz 3,746 Reputation points

2023-05-27T07:05:41.4166667+00:00

This is a duplicate of https://learn.microsoft.com/en-us/answers/questions/1290668/why-array-variable-doesnt-work-inside-the-if-state which has been answered.
Debojit Acharjee 455 Reputation points

2023-05-28T06:44:55.4066667+00:00

Which was the right answer do you think?

Barry Schwarz 3,746

Let's try again.

There are three arrays named a.

The the one that you define in the if statement is created when the definition is encountered and is destroyed when you exit the compound statement at the }.  It is visible only to code that executes in the if statement.

The one defined after the if statement is created when your execution reaches that point and is destroyed when main terminates. It is visible to all the code in main that follows

The one defined at the top of your code is created before your program executes and is destroyed when your program terminates.  It is visible everywhere in your code EXCEPT for the areas of your code where the other two are visible.  Each of the other two definitions temporarily hides this one from view.

Given this information, you should be able to figure out which value each of the three arrays are initialized with and which value is the only one visible at the time you call printf.

Debojit Acharjee 455 Reputation points

2023-05-29T03:58:01.2833333+00:00
Was that an answer or a mystery?

I have the answer for you--this was not working because we can't declare any variable inside a conditional statement.

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c=0; if (b < 0) { int c = 1; } printf("%d, %s", c, a); return 0; }

In this program, the value of 'c' doesn't change to '1' because it was declared as "int" inside the if-statement.
Barry Schwarz 3,746 Reputation points

2023-05-29T08:52:03.6033333+00:00

You are just flat wrong. You can define variables inside the compound statement that follows an if clause. Your code does just that. What you cannot do is reference those variables once you leave the compound statement.

There are two variables named c. The second one is initialized to 1 just as you coded it. The reason printf cannot see it is because it is destroyed when you exit the if statement. Even if it had not been destroyed, it still would not be visible to printf. You need to review the rules regarding scope.
Debojit Acharjee 455 Reputation points

2023-05-30T06:07:02.5866667+00:00
I don't think declaration is possible inside the if-statement. IN the following program I have removed the declaration inside the if-statement but only defined the 'c' variable. Now it can print the new value of 'c'.

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c=0; if (b < 0) { c = 1; } printf("%d, %s", c, a); return 0; }

Output:

1, RED

This was not possible when the variable 'c' is declared as "int" inside the if-statement. Do you have any answer for that?
YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-06-01T01:33:26.5633333+00:00

Hi Debojit Acharjee,

May I know if you have got any chance to check my answer? If the reply is helpful, please click "Accept Answer" and kindly upvote it. It will also help others to solve the similar issue.

1 answer

Your answer

Barry Schwarz 3,746 Reputation points

2023-05-27T07:05:41.4166667+00:00

This is a duplicate of https://learn.microsoft.com/en-us/answers/questions/1290668/why-array-variable-doesnt-work-inside-the-if-state which has been answered.
Debojit Acharjee 455 Reputation points

2023-05-28T06:44:55.4066667+00:00

Which was the right answer do you think?
Barry Schwarz 3,746 Reputation points

2023-05-28T13:35:53.33+00:00

Let's try again.

There are three arrays named a.

The the one that you define in the if statement is created when the definition is encountered and is destroyed when you exit the compound statement at the }. It is visible only to code that executes in the if statement.

The one defined after the if statement is created when your execution reaches that point and is destroyed when main terminates. It is visible to all the code in main that follows

The one defined at the top of your code is created before your program executes and is destroyed when your program terminates. It is visible everywhere in your code EXCEPT for the areas of your code where the other two are visible. Each of the other two definitions temporarily hides this one from view.

Given this information, you should be able to figure out which value each of the three arrays are initialized with and which value is the only one visible at the time you call printf.
Debojit Acharjee 455 Reputation points

2023-05-29T03:58:01.2833333+00:00

Was that an answer or a mystery?

I have the answer for you--this was not working because we can't declare any variable inside a conditional statement.

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c=0; if (b < 0) { int c = 1; } printf("%d, %s", c, a); return 0; }

In this program, the value of 'c' doesn't change to '1' because it was declared as "int" inside the if-statement.
Barry Schwarz 3,746 Reputation points

2023-05-29T08:52:03.6033333+00:00

You are just flat wrong. You can define variables inside the compound statement that follows an if clause. Your code does just that. What you cannot do is reference those variables once you leave the compound statement.

There are two variables named c. The second one is initialized to 1 just as you coded it. The reason printf cannot see it is because it is destroyed when you exit the if statement. Even if it had not been destroyed, it still would not be visible to printf. You need to review the rules regarding scope.
Debojit Acharjee 455 Reputation points

2023-05-30T06:07:02.5866667+00:00

I don't think declaration is possible inside the if-statement. IN the following program I have removed the declaration inside the if-statement but only defined the 'c' variable. Now it can print the new value of 'c'.

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c=0; if (b < 0) { c = 1; } printf("%d, %s", c, a); return 0; }

Output:

1, RED

This was not possible when the variable 'c' is declared as "int" inside the if-statement. Do you have any answer for that?
YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-06-01T01:33:26.5633333+00:00

Hi Debojit Acharjee,

May I know if you have got any chance to check my answer? If the reply is helpful, please click "Accept Answer" and kindly upvote it. It will also help others to solve the similar issue.

Answer 1

YujianYao-MSFT 4,296 Microsoft External Staff

Hi Debojit Acharjee,

In your code, a is a global variable that is initialized to "RED" at the beginning of the program. And in the main function, you declare a local variable a and assign it the value "GREEN". However, when you refer to the variable a in the printf statement, it still refers to the global variable a.

This is because in C language, local variables override global variables of the same name. The local variable a declared in the if statement block will shadow the scope of the global variable a, but it is only valid inside the if statement block. When you refer to a in the printf statement, it refers to the global variable a, not the local variable. And c is not redefined, it is still previously defined, so its value will change.

If you set a breakpoint, we could see that the value of a in the if block will become "GREEN", and the value of a will no longer be "GREEN" when the if block is run out.

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Best regards,

Elya Yao

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YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-05-30T07:06:30.0933333+00:00
Hi Debojit Acharjee,

Only one c is defined in the following code, so you could modify its value. If you use int c in the if block, you could only modify the c in the if statement block, not the c in int b=-1, c=0;.

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c=0; if (b < 0) { c = 1; } printf("%d, %s", c, a); return 0; }
YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-06-01T05:26:33.5933333+00:00

Hi Debojit Acharjee,

May I know if you have got any chance to check my answer? If the reply is helpful, please click "Accept Answer" and kindly upvote it. It will also help others to solve the similar issue.

Debojit Acharjee 455

Inside the main function it's possible to declare a variable many times, but why not inside the if?

Even when declaring the c only in the if statement the program won't work because it gives error.

#include <stdio.h>

char a[10] = "RED";

int main ()
{

int b=-1;

if (b < 0)
{

  int c = 1;

}

    printf("%d, %s", c, a);

    return 0;
}

Output:

tempCodeRunnerFile.c:17:22: error: 'c' undeclared (first use in this function)
     printf("%d, %s", c, a);
                      ^
tempCodeRunnerFile.c:17:22: note: each undeclared identifier is reported only once for each function it appears in

Barry Schwarz 3,746 Reputation points

2023-06-03T19:40:36.99+00:00

This has been explained before.

The variable c that you declare inside the if statement no longer exists once you exit that statement. Therefore, there is no variable c when you call printf.

If you move the printf statement inside the brackets, you will get the result you expect.
Debojit Acharjee 455 Reputation points

2023-06-04T13:28:16.4833333+00:00
That's not the case because c works only when it's not declared but defined inside the if statement. I think we can't declare any variable inside conditional statements in C.

This program works:

#include <stdio.h> char a[10] = "RED"; int main () { int b=-1, c; if (b < 0) { c = 1; } printf("%d, %s", c, a); return 0; }

Output:

1, RED
Barry Schwarz 3,746 Reputation points

2023-06-04T18:29:33.9233333+00:00

You can think what you like but the fact is that in your latest code c is not defined inside the if statement. It is defined in the statement before the if, along with b. It is assigned a value inside the if. That assignment does not affect the life or visibility of c. Once c is assigned a value, it will keep that value until you change the value or c gets destroyed. Neither happens before you call printf.

If you change the initialization value of b to a non-negative number, the assignment inside the if will not execute and the value of c when you attempt to print it will be indeterminate.
Debojit Acharjee 455 Reputation points

2023-06-05T06:02:57.8833333+00:00

What do you mean by defining? I haven't defined c outside if statement but declared it with the b as integer. I have defined c = 1; inside the if statement in the above program. But when I declare the c as int inside the if statement, it doesn't work.
YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-06-05T07:22:28.5333333+00:00
Hi Debojit Acharjee,

In this code, you define a c in the main function.

int b=-1, c;

The c in the if block is using the previously defined variable(int b=-1, c;).

if (b < 0) { c = 1; }

So the program outputs 1, RED.
Barry Schwarz 3,746 Reputation points

2023-06-05T10:19:11.5033333+00:00

The statement

c = 1;

is not a definition. It is an assignment statement and only has meaning because c was previously defined. That definition occurred in the declaration

int b=-1, c;

which was processed earlier in the program and before the if statement.

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Why array is not initialized inside if-statement?

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