Member Avatar for boyz

1.main()
2.{
3. int a=0;
4. while(...)
5. {
6. int a=9;
7. }
8.}

now friendz question is
question is what make a of 3 global and a of line 6 local what is happend at the time of local decla.. and at the time of global decl...

would u plz help me to understand this...

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Short answer: neither a in the snippets is global. The outer a is local to main (it has block scope for the main function); the inner a is local to the loop block and shadows the outer name while that block is active. The inner variable is created and initialized each time execution reaches its declaration and is destroyed when the block exits; the outer variable keeps its value across the block boundary.

was right to say these are not global, and is correct about the inner declaration taking precedence inside the loop. If observed the outer value changing after the loop, the usual causes are: the inner declaration was accidentally omitted (so the code was assigning to the outer variable), braces/scoping were different than expected, or the test prints were placed where they showed the assigned value. To debug, print the value before, inside, and after the block and compile with warnings that catch shadowing (for GCC/Clang use -Wshadow).

A minimal reproducible test:

#include <stdio.h>

int main(void) {
    int x = 1;
    printf("outer before: %d\n", x);
    for (int i = 0; i < 1; ++i) {
        int x = 2; /* shadows outer x */
        printf("inside: %d\n", x);
    }
    printf("outer after: %d\n", x);
    return 0;
}

Compile with gcc -std=c99 -Wall -Wextra -Wshadow test.c -o test and run; expected output shows inside different from outer after. For authoritative details on scope and storage duration see C: scope and C: storage duration.

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All 9 Replies

Member Avatar for doublex

1.main()
2.{
3. int a=0;
4. while(...)
5. {
6. int a=9;
7. }
8.}

now friendz question is
question is what make a of 3 global and a of line 6 local what is happend at the time of local decla.. and at the time of global decl...

would u plz help me to understand this...

hi there,
i don't know if i understood your question...
if you want to know how those variables are stored in memory
well, in my opinion, none of them are global. the one declared in line 3 is local to main function, and resides on the respective stack, and the one declared in the while loop, should be put in a register, because it is accessed very often, and the compiler can see this.

Member Avatar for thinker

1.main()
2.{
3. int a=0;
4. while(...)
5. {
6. int a=9;
7. }
8.}

now friendz question is
question is what make a of 3 global and a of line 6 local what is happend at the time of local decla.. and at the time of global decl...

would u plz help me to understand this...

i think a in 3rd line is not global its local to the main function and if you change the value of a in the 6th line or ne where inside the program it will display the new value...

Member Avatar for Jishnu

Use code tags :D

Member Avatar for Jishnu

i think a in 3rd line is not global its local to the main function and if you change the value of a in the 6th line or ne where inside the program it will display the new value...

ABSOLUTELY WRONG !!!

Member Avatar for Jishnu

a on line 3 is local to main(), while the scope of a in line 6 is limited to the while loop. Inside the while loop, the most local variable is given preference. And if you change a inside the while loop, the value of a on line 3 will remain unaffected. Try it out by using printf statements in between.

Member Avatar for doublex

jishnu is right :D

Member Avatar for thinker

i tried it but not working as u r saying...!!!!!!!!!

Member Avatar for thinker

but the updation is done in the while loop and so the value of a will be the new value even if we print the value after the termination of the loop.......:-/

Member Avatar for Salem

How about posting your latest code instead of just moaning it doesn't work.

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