6 replies to this topic

[keithwb]

tried to readup on the linkage, i noe there is internal linkage, external linkage, no linkage. but my understanding is still not enough to solve this question.

someone pls enlighten me, and hopefully with explanation.

thank in advance!


which of the four given statements is true about the following program?

int i = 0;

void f(long i);

void main (void){

	f(i);

}

void f (long i){

	long k = i;

	static int j = 1;

	if(j>1)

		k = i+2;

	j = j*2;

}

(a) variable i at line 1 has interenal linkage
(b) variable i at line 6 has function prototype scope
© variable j at line 8 has automatic storage
(d) varirable k at line 7 has no linkage

[dargueta]

This screams "homework." We gladly help with homework, but we don't do it for you. Do you know what the difference between linkage and scope is?

[keithwb]

i read up on linkage, in fact i see no much difference between linkage and scope.

for example, function prototype scope is same as internal linkage?

global scope is as external linkage?

i pretty confused.

[dargueta]

No, but you're on the right track.

#include <bar.h>

void foo(int a, int b);
extern void bar(void);

int main(int argc, char **argv)
{
    foo(argc, argc+1);
    bar();
    return 0;
}

void foo(int a, int b)
{
    ... do something ...
}

Take a guess.

Edited by dargueta, 16 March 2010 - 12:01 AM.
Forgot code

[keithwb]

dargueta said:

No, but you're on the right track.

#include <bar.h>

void foo(int a, int b);
extern void bar(void);

int main(int argc, char **argv)
{
    foo(argc, argc+1);
    bar();
    return 0;
}

void foo(int a, int b)
{
    ... do something ...
}

Take a guess.

try to print out the value of a and b in foo? sorry if that a noob answer, as i pretty new in this.

[dargueta]

No, I meant try to guess the linkage of the functions, and the scope of the variables.

[keithwb]

dargueta said:

No, but you're on the right track.

#include <bar.h>


void foo(int a, int b);

extern void bar(void);


int main(int argc, char **argv)

{

    foo(argc, argc+1);

    bar();

    return 0;

}


void foo(int a, int b)

{

    ... do something ...

}

Take a guess.

oh ok,

bar() is global scope, do not necessary have to put extern in front. it will still work.

foo() is global scope as well.

argc is external linkage?

int a and b is function protype scope.