Expression must be modifiable lvalue

Question

typedef struct state1_s {
    u8 size;
    u8 state;
} state1_t;
typedef struct state2_s {
    u8 size;
    u8 state[2];
} state2_t;
typedef struct state3_s {
    u8 code;
    u8 count;
}state3_t;

I have these three structures. When i do the following

state1_t comp[8];
state2_t *avail;
state3_t *health;

&comp = (state1_t *)(&(avail->state[2])+1);

i get a expression must be modifiable lvalue error.

I get the same error when i do

&comp = (state1_t *)(&(heaalth->count) +1);

how do i fix this? I want the address of comp to be at the end of one structure. How do i do that?

You can't change the address of a variable. – Barmar Sep 10 '14 at 16:50 — Barmar, Sep 10 '14 at 16:50

score 0 · Answer 1 · answered Sep 10 '14 at 16:54

0

state1_t comp[8];

In this declaration comp is an array of 8 struct's state1_t with base address of the array being comp which cannot be modifiable. Hence the modifiable lvalue error when you try changing address of comp.

answered Sep 10 '14 at 16:54

Sunil Bojanapally

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score 0 · Answer 2 · answered Sep 10 '14 at 16:54

You can't change the address of an array. You should use a pointer instead:

state1_t *comp;

Then you can do:

comp = (state1_t *)(&(avail->state[2]));

You don't need to add 1 to the address -- state[2] is already the end of the structure, since state[1] is the last element of that array.

Expression must be modifiable lvalue

2 Answers2