Hi ALL, When I run the follwoing code, I get the follwoing 1245048 1245044 1245052 6 I tried changing the value of 'a' and the last line of the code, the printf prints whatever is assigned to 'a'. Please explain me this behaviour. Code: void main() { int a=6; int b=1024; int *swap=(&b-1); printf("%u\n",&b); printf("%u\n",swap); swap+=2; printf("%u\n",swap); printf("%d\n",*swap); } Regards, Swapna
You are printing the address of b which is in your system 1245048 and swap which is 4 bytes less which should be 1245044 and swap after adding 2 should be 1245044 + 2 = 1245052 which is what is expected.
Yes u r right.. then what is *swap in the last line??? The value present in 1245052 at that point of time.. right??? If u can run the program in ur system.. u can find that, the value is nothing but 'a', in this case 6. If u change the value of 'a' then the output also changes.. Any idea fro this type of behaviour??
swap is set to the address of the integer before b (&b minus 1), in this case, a. Therefore, when you print what swap points to (a), you get the contents (a). This only works because your machine happens to have a common implementation of the stack. There is no requirement of the language that requires it to be so.
Mr DaWei Let us consider the follwoing example: &a=1245044 &b=1245048 Now, swap=&b-1, swap=1245044(According to ur explanation) swap+=2, swap=swap+2, swap=1245052(if int occupies 4 bytes) *swap= contents(1245052), swap= unknown value not 'a'. Please correct me if i am wrong Regards, Swapna
Quite simple, should be no confusion. Just pointer manipulation. Understand whats actually procedure of pointer storage and you will make it. Or take a K&R and make your task simple.....
It's almost certainly the case that your system is implemented in such a way that locals and automatic variables are allocated on the stack, and this in a system, such as the x86, where the stack grows downward. Presuming that 'a' is located at 1245412, in your case, then the variables are at the following addresses: Code: 1245412 a 6 1245408 b 1024 1245404 swap 1245404 (&b - 1*sizeof int) swap then becomes, after += 2*sizeof int, 1245404 + 8, 1245412, which is the address of a when dereferenced, then, the value of a (6) is printed. Your number, 1245052, seems to be a typo. Again, this is not only implementation dependent, hardware-wise, but also, software-wise. Exercises like this are only useful when produced on the same system with the same OS and the same compiler. Just a caveat.
