Const Int and Points

[FIG-GHD742]

Hello.

I have a question about Const Int and Points

I have this code

const int j=18;
    int *k= (int)&j;
    *k=110;
    printf("Data: %d -> %d; %d : %d-> %d\n", &j, j, k, &k, *k);

this well return for me
Data: 2280676 -> 18; 2280676 : 2280672-> 110
but if i remove "const" well i have
Data: 2280676 -> 110; 2280676 : 2280672-> 110 this work.

But how can same memory address have two int ( 18 and 110 )
Can somebody tell me how this are possible?

[ankur_]

Hello.

I have a question about Const Int and Points

I have this code

const int j=18;
    int *k= (int)&j;
    *k=110;
    printf("Data: %d -> %d; %d : %d-> %d\n", &j, j, k, &k, *k);

this well return for me
Data: 2280676 -> 18; 2280676 : 2280672-> 110
but if i remove "const" well i have
Data: 2280676 -> 110; 2280676 : 2280672-> 110 this work.

But how can same memory address have two int ( 18 and 110 )
Can somebody tell me how this are possible?

Try again it will give 110 for both

[sbesch]

I'm not 100% sure about this, but I suspect that you are seeing the result of a compiler optimization. The "const" is telling the compiler that the value of j won't be changing. On the other hand, you bypass that assumption by trying to sneak a new value into j's memory address. j's address does indeed contain 110 at run time, but when the print statement comes along, the compiler assumes that j is still 18 (that's what you told it to assume with the const) and doesn't bother to reference j's memory location at all - it just loads 18 using an immediate operand with a value of 18.

The immediate reference may need some clarification if you've never coded in assembly. There are at least 2 ways to load numeric values. The first, and most versatile, is to use an assembler instruction that loads a value from memory. In this case, the address of the value is embedded in the machine code for the instruction that loads the value. In most cases this will be a 32 bit pointer. In the second case, the value itself is coded directly into the machine code. The length of this so called immediate operand may be 1, 2, 4, or 8 bytes. Immediate loads are almost always more efficient, both in terms of memory utilization and speed. In your case, the compiler will be able to code the constant in 1 byte (and then sign extend it to 32 bits) and the job is done as soon as the instruction is loaded. A memory reference would require a 32 bit address load (3 extra bytes) followed by the operand (also 32 bits). In total 7 extra bytes to load.

[FIG-GHD742]

Thank you sbesch. I think i understand what you men.
I have this question to day in school and my teacher are not sure.

Can i edit or mark this thread as finhis?

/ Fig

[Aia]

Can i edit or mark this thread as finhis?

When you are the original poster you'd see at the bottom, after the last post, the mark as solved title, in blue. Click on it.