绘制执行代码图

Question

Consider the following lines of code:

char a = 'a';
char b = 'b';
int ai[] = { 1, 2 };
int i = 0;

Assume that word size is 32 bits, that an int is 32 bits and that memory allocations are made in the reverse order to the declarations starting at address location 68.

Draw a diagram that shows the effect of executing the following lines of code.

 for (i = 0; i < 8; i++)
 *(&a – i) = 'z';

So I'm pretty unsure what it's expecting of me here. My initial guess was it loops through and essentially does a-1,a-2....,a-8 making each value equal to the character Z. So the diagram would essentially be 8 objects all pointing to Z each having the same address as Z. I'm not sure if that's right or horribly wrong. Could anyone care to explain. FYI this is a past exam paper, practising for finals, not examined work!

Answer 1

You can't guarantee how this works as it's up to the C compiler how it lays things out. However, you can easily establish how a particular C compiler does it. Assuming gcc , compile with -g , then run the program under gdb . Use print to print the variables (or print &x to print the address of a variable), and x to examine memory.

Answer 2

The code is pretty strange as it makes no sense. As far as I can tell from your information this will happen:

i will be located at address 68

ai will be located at 72 and 76

The difficult part is a and b. They only requires a byte but the compiler may still word align them! It depends on your system/compiler.

Option 1 using word alignment:

b will be at address 80

a will be at address 84

Option 2 NOT using word alignment:

b will be at address 80

a will be at address 81

The loop will start by overwriting a with az Then it will decrement the address by one and overwrite that location with z

This will repeat 8 times.

So for option 1 the result will be that both a and b will be z and some unused memory will also be z.

memory before:

00 00 00 00

00 00 00 01

00 00 00 02

-- -- -- 'b'

-- -- -- 'a'

memory after:

00 00 00 08

00 00 00 01

00 00 00 02

'z' 'z' 'z' 'z'

'z' 'z' 'z' 'z'

For option 2 both a and b will be z but also ai will be changed. ai[1] = int(zzzz) and ai[0]=int(zz)

memory before:

00 00 00 00

00 00 00 01

00 00 00 02

'b' 'a'

memory after:

00 00 00 08

00 00 'z' 'z'

'z' 'z' 'z' 'z'

'z' 'z'

EDIT:

I see my answer got down voted.

Maybe because I forgot to tell that it was for big endian systems.

So I tried the program on a windows machine as well and got this result:

a address 28fec3

b address 28fec2

ai address 28feb8

i address 28feb4

BEFORE

0 0 0 0 (i=0)

1 0 0 0 (ai[0]=1)

2 0 0 0 (ai[1]=2)

-43 -116 ba (unused unused b='b' a='a')

AFTER

8 0 0 0 (i=8)

1 0 0 0 (ai[0]=1)

zzzz (ai[1]=int("zzzz")

zzzz (unused unused b='z' a='z')

Due to little endian things "turned around" but the main point is the same.

Thanks for the down vote with no comment.