Place A Variables Address Into Another Variable In C

Question

I was wondering if I could capture a variables address that I could display by using

fprintf(stdout, "%p\n", (void*) &var);

and actually place it into another variable, without displaying it.

(Into a String variable, not into it's actual address)

I have tested this with setbuf and everything but it still displays the String which I would not want.

I have tried looking everywhere but all I see is using either setbuf or some other means that doesn't fit what I am looking for.

So you want to place the pointer's address as a string into a char *? Have you tried sprintf(mystr, "%p", (void*) &var)? Here mystr would be a pointer to a character array. — Squirrel
– Squirrel, Commented Apr 30, 2015 at 18:42

Sergey Kalinichenko · Accepted Answer · 2015-04-30 18:51:16Z

4

You can capture a string representation of an address in the same way that you capture any output in a string - by using sprintf function:

char buf[17];
sprintf(buf, "%p", (void*) &var);

At this point buf contains a null-terminated string with a representation of var's address suitable for printing.

Of course if you wish to keep the string beyond the scope of buf you need to allocate another string dynamically, and strcpy your buf into it.

edited Apr 30, 2015 at 18:51

answered Apr 30, 2015 at 18:43

Sergey Kalinichenko

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1 Comment

jarmod Over a year ago

Yes, though probably minus the \n.

Iłya Bursov · Accepted Answer · 2015-04-30 18:42:09Z

2

You can use something like this:

void* pointerToVar = &var;

fprintf(stdout, "%p\n", pointerToVar);

answered Apr 30, 2015 at 18:42

Iłya Bursov

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Comments

Kyle · Accepted Answer · 2015-04-30 18:59:53Z

0

I tried the first post and it worked also.

We also tried a modified version of this and it also works like this.

char a = 5; // Random number, literally has no importance

unsigned long long int b = (unsigned long long int) &a;

int num = abs(b);

answered Apr 30, 2015 at 18:59

Kyle

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1 Comment

user3629249 Over a year ago

abs() requires an int parameter, not a unsigned long long int parameter. using the suggested code causes the compiler to raise a warning. and the value is already unsigned, so why bother with abs()?

scoutcat · Accepted Answer · 2015-04-30 18:47:05Z

-1

pointer is really just an integer (address). You could do this:

float var;
int addr_of_var = (int)(&var);

depending on the platform, it might be long int, I think.

answered Apr 30, 2015 at 18:47

scoutcat

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2 Comments

alk Over a year ago

The won't work on most 64bit systems. Use intptr_t or uintptr_t instead of int or just void* for any pointer.

alk Over a year ago

"pointer is really just an integer" this might be true for some implementations, but definitly not for all. For an example you might like to read here: en.wikipedia.org/wiki/Memory_segmentation (It's been not so long we used this... ;-))

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