TypeDef Struct Variable to different #define statement during variable init.

the TI demo code provided the statement (without error). I need help to understand and fix this error.

RFFT32 fft=RFFT32_512P_DEFAULTS; (There are 6 different default setting).

I wish to implement switch statement which setup the fft to different setting

RFFT32 fft;

case 0:
fft = RFFT32_512P_DEFAULTS;
break;
case 1:
fft = RFFT32_256P_DEFAULTS;
break;
case 2:
fft = RFFT32_128P_DEFAULTS;
break;

However the compiler report error "Error expected a expression".

I have tried fft = (RFFT32 ) RFFT32_128P_DEFAULTS; but no luck as it report error (RFFT32) is not allowed

The compiler is Code Composer V6 (from TI) which is based on Eclipse.

There are 6 different default based on RFFT32_512P_DEFAULTS
RFFT32_512P_DEFAULTS
RFFT32_256P_DEFAULTS
RFFT32_128P_DEFAULTS, etc

where RFFT32_512P_DEFAULTS has the following code provided by TI example code.
//------------------------------------------------------------------------------------
#define RFFT32_512P_DEFAULTS { (long *)NULL,\
(long *)NULL,\
512,\
9,\
(long *)NULL,\
(long *)NULL,\
0,\
0,\
8,\
(void (*)(void *))FFT32_init,\
(void (*)(void *))FFT32_calc,\
(void (*)(void *))FFT32_mag,\
(void (*)(void *))FFT32_win
}
//-------------------------------------------------------------------------
The RFFT32 typedef struct has the following statement

typedef struct {
long *ipcbptr;
long *tfptr;
int size;
int nrstage;
long *magptr;
long *winptr;
long peakmag;
int peakfrq;
int ratio;
void (*init)(void *);
void (*calc)(void *);
void (*mag)(void *);
void (*win)(void *);
}RFFT32;
typedef RFFT32 *RFFT32_handle;

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LVL 85

ozoCommented: 2015-03-12

There should be a \ at the end of
(void (*)(void *))FFT32_win

#define RFFT32_512P_DEFAULTS { (long *)NULL,\
(long *)NULL,\
512,\
9,\
(long *)NULL,\
(long *)NULL,\
0,\
0,\
8,\
(void (*)(void *))FFT32_init,\
(void (*)(void *))FFT32_calc,\
(void (*)(void *))FFT32_mag,\
(void (*)(void *))FFT32_win\
}

Richard PayneChief Technology EngineerAuthor Commented: 2015-03-12

Thank, I actually put the '}' there to make it clearer where it suppose to be

(void (*)(void *))FFT32_win}

as original code.

This does not resolve problem.....Anyone?

switch(0){
case 0:
{RFFT32 f=RFFT32_512P_DEFAULTS;
fft = f;
break;}
}

LVL 36

sarabandeCommented: 2015-03-12

to add to ozo's solution:

the point is that you only can initialize a struct with an initializer block {} but not assign a block to a variable.

so you could initialize helpers and assign the helper variables :

RFFT32  fft;
RFFT32  fft0 = RFFT32_512P_DEFAULTS;
RFFT32  fft1 = RFFT32_256P_DEFAULTS;
....

 case 0:
     fft = fft0;
    break;
 case 1:
     fft = fft1;
     ...

Open in new window

Sara

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TypeDef Struct Variable to different #define statement during variable init.

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