CCS uint

user4451984

Other Parts Discussed in Thread: MSP430FR5969

Hi, I'm writing a code on CCS for a MSP430FR5969 and I don't understand why it doesn't work as I want.

This is the code, and when the last value of the array (1280) is selected the X_Time is something around 64848 (it seems a problem with the size of the int) instead of 128000. I have changed the array from uint16_t to uint32_t and all works well.

uint32_t X_Time=0;
const uint16_t TABLE_X_TIME_MODE_2_0to3[16] = {5,10,15,20,30,40,60,80,120,160,240,320,480,640,960,1280};
X_Time=TABLE_X_TIME_MODE_2_0to3[POT_Frq_0_15]*100;

So I review the code and I find another array :

uint16_t Gap=0;
const uint8_t TABLE_MODE_0[8][2] ={{1,1},{2,1},{3,1},{4,1},{1,5},{2,5},{3,5},{4,5}};
Gap=TABLE_MODE_0[Program][1] *1000;

In this case all works well without change the array from uint8_t to uint16_t.

Anyone know why it happens ?

Thanks

Riccardo

over 9 years ago

0 Clemens Ladisch over 9 years ago

Guru 297970 points

128000, in binary, is 11111010000000000. That's more than 16 bits.

0 user4451984 over 9 years ago in reply to Clemens Ladisch

Intellectual 850 points

I don't understand,the value in my array is 1280 not 128000, for the same reason I should have an error with the uint_8 array

0 Clemens Ladisch over 9 years ago in reply to user4451984

Guru 297970 points

As specified in section 6.3.1.8 of the C standard, arithmetic operations (like the multiplication) use type "int" or "unsigned int" unless you are explicitly using larger types. On the MSP430, these types have 16 bits.

0 user4451984 over 9 years ago in reply to Clemens Ladisch

Intellectual 850 points

So if the result is higher than 65585 I have to use at least one uint32_t operator even if my X_Time variable is a uint32_t, isn't it ?
If I use 100 instead of 1000 in the second case it should work because the compiler use 16 bit to represent 100 ?

0 Clemens Ladisch over 9 years ago in reply to user4451984

Guru 297970 points

Yes, the multiplication is independent of the assignment. To get a 32-bit multiplication, at least one of the operands must have 32 bits, or be explicitly cast to a 32-bit type:

uint32_t Gap = TABLE_MODE_0[Program][1] * (uint32_t)1000;

Both 100 and 1000 would be represented as 16 bits. The only difference is whether the result of the multiplication would fit into 16 bits.

0 user4451984 over 9 years ago in reply to Clemens Ladisch

Intellectual 850 points

Thanks !

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CCS uint