expected expression before '{' token

Question

I am getting: "error: expected expression before '{' token" for the line I've commented before. If the struct is already defined why would it need a "{" before token. Thanks for any help you can provide.

struct sdram_timing {
    u32 wrdtr;
    u32 clktr;
};

int calibration(void);
unsigned char read_i2c_cal(void);
static unsigned int eepcal[15];

main() {
    DQS_autocalibration();
}

int calibration(void)
{
    struct sdram_timing scan_list[30];

    read_i2c_cal();
    if(eepcal[0] == 0){

       scan_list = {{eepcal[1], eepcal[2]}, {-1, -1}}; // <-- PROBLEM LINE

        }
        else {
            //foo
        }

    return 0;
}

unsigned char read_i2c_cal(void) {
    eepcal[0] = 0;
    eepcal[1] = 02;
    eepcal[2] = 03;
}

Incest

Answer 1

The error is because you can't assign an array that way, that only works to initialize it.

int arr[4] = {0}; // this works
int arr2[4];

arr2 = {0};// this doesn't and will cause an error

arr2[0] = 0; // that's OK
memset(arr2, 0, 4*sizeof(int)); // that is too

So applying this to your specific example:

struct sdram_timing scan_list[30];
scan_list[0].wrdtr = 0;
scan_list[0].clktr = 0;

or you could use memset the same way, but instead of sizeof(int) you need size of your structure. That doesn't always work... but given your structure, it will.

Lotze

so after it is initialized using "struct sdram_timing scan_list[30];" how can I assign values to scan_list? – Incest

@Incest - What I gave you was the generic way to do it. I edited my post to give you a specific example using your code. – Lotze

This also happens when trying to assign a struct (e.g. struct { int a; int b; } foo; foo = {1, 2};)... +1 for differentiating assignment and initialization in a way that gave me a quick answer. – Tuberose

Answer 2

Arrays in C language are not assignable. You can't assign anything to the entire array, regardless of what syntax you use. In other words, this

scan_list = { { eepcal[1], eepcal[2] }, {-1, -1} };

is not possible.

In C89/90 you'd have to spell out your assignments line by line

scan_list[0].wrdtr = eepcal[1];
scan_list[0].clktr = eepcal[2];
scan_list[1].wrdtr = -1;
scan_list[1].clktr = -1;

In modern C (post-C99) you can use compound literals to assign entire structs

scan_list[0] = (struct sdram_timing) { eepcal[1], eepcal[2] };
scan_list[1] = (struct sdram_timing) { -1, -1 };

Finally, in modern C you can use memcpy and compound literals to copy data to the array

memcpy(scan_list, (struct sdram_timing[]) { { eepcal[1], eepcal[2] }, {-1, -1} },
  2 * sizeof *scan_list);

The last variant, albeit not very elegant, is the closest way to "emulate" array assignment.

Melano

Answer 3

You can only use an initializer list in the declaration of the variable, not after the fact.

Kumquat

Answer 4

Initializer list can only be used to initialize an array. You cannot use it afterwards.

However if you use GCC, you can use Compound Literal extension:

scan_list = (struct sdram_timing[30]){{eepcal[1], eepcal[2]}, {-1, -1}};

You might need to change scan_list type to be struct sdram_timing *

Loudermilk

Compound literals do exist in C99, but that doesn't mean that arrays are assignable in C99. They are not. Your code will not compile. – Melano

what if I want to modify values of scan_list after it is initialized? – Incest

@AndreyT I just realized. Array variables cannot stand alone as a lvalue. Need to change to a pointer – Loudermilk

@Incest the code above is showing you how to change it after initialization. – Loudermilk