如何将整数转换为uint8_t指针？

Question

I have a function defined like:

void foo(uint8_t *val, int num)
{
    for(int i=0;i<num;i++)
        printf("Buffer[%i]=0x%X \n",i,val[i]);
}

As a global variable, I have an array declared as:

uint8_t Buffer[4] = {0x01, 0x02, 0x03, 0x04};

So in order to print the buffer, I do the following:

int main()
{
    foo(Buffer,4);
    return 0;
}

Which gives as a result:

Buffer[0]=0x1                                                                                    
Buffer[1]=0x2                                                                                               
Buffer[2]=0x3                                                                                               
Buffer[3]=0x4 

The thing is that, for a particular case, I need to send only one uint8_t parameter to that function replacing the buffer, so I implemented it like:

int main()
{
    uint8_t READ_VAL[] = {0x01};
    foo(READ_VAL,1);
    return 0;
}

Anyway, is there any way to do it inline? I tried to do it like

foo((uint8_t *)0x01,1);

but it is not working (gives me Segmentation fault error). Any idea how can I do it?

Answer 1

foo((uint8_t []) { 0x01 }, 1);

The form “ ( type ) { initializers … } ” is a compound literal , specified in C 2018 6.5.2.5. It creates an object of the specified type with the value given by the initializers.

(uint8_t []) { 0x01 } creates an array of one uint8_t with value 0x01 . As an array, it is automatically converted to a pointer to its first element, which is suitable for the first parameter of foo .

A compound literal inside a function is temporary, with automatic storage duration associated with its enclosing block. A compound literal outside a function endures for the execution of the program, with static storage duration.