Can't pass and array??

[see the big C]

Hello,

here's some code:

uint8_t receiveBuffer[256];  // Array filled elsewhere! 



void abc(uint8_t *receiveBuffer[]){
uint8_t a[256];

strncpy(a,receiveBuffer, 50);   //<<< a has junk contents????
}




int main(void)
{
    abc(receiveBuffer);

}

Why can't a array contain the contents of receiveBuffer???

Thank you all for your help

[laserlight]

This declares abc as a function with void return type that has a parameter of type uint8_t**:

void abc(uint8_t *receiveBuffer[]){

Notice the double asterisks, i.e., receiveBuffer is declared to be a pointer to a pointer to uint8_t. Now, when you pass an array of 256 uint8_t as an argument, it is converted to a pointer to its first element. Therefore, you probably want:

void abc(uint8_t receiveBuffer[]){

or equivalently:

void abc(uint8_t *receiveBuffer){

Note that this kind of type mismatch is something that your compiler can warn you about, so compile at a high warning level and pay attention to compiler warnings.

[jimblumberg]

Why can't a array contain the contents of receiveBuffer???

First strncpy() requires the source "array" to be a valid C-string. Are you sure receiveBuffer is a properly terminated C-string?

Second your receiveBuffer may contain upto 256 characters yet you're only trying to copy 50 characters. Remember that strncpy() doesn't automatically add the end of string character, you need to manually insure that the end of string character exists and add it if it doesn't.

[see the big C]

Hi,

Okay, it works .... I shouldn't be using strcpy since they are numbers
so I simply accessed the array by an equals sign see code below.

uint8_t receiveBuffer[256];


void abc(uint8_t receiveBuffer[]){
uint8_t a[256];

//strncpy(a,receiveBuffer, 50);

a[12] = receiveBuffer[12];
}



int main(void)
{
   abc(receiveBuffer);

}

Its been 5 years I stopped C... LOL ... I would need a good refresher LOL

Thanks so much for your help!
Bob

[Salem]

Try memcpy() then.

[see the big C]

Yes thank you Salem
StbC