CCS C Compiler Manual PCB / PCM / PCH

Common Questions & Answers
}
set_tris_b(ALL_IN);
i = PORTB;
// i now contains the portb value.
Remember when using the #BYTE, the created variable is treated like memory. You must
maintain the tri-state control registers yourself via the SET_TRIS_X function. Following is an
example of placing a structure on an I/O port:
struct port_b_layout
{int data : 4;
int rw : 1;
int cd : 1;
int enable : 1;
int reset : 1; };
struct port_b_layout port_b;
#byte port_b = 6
struct port_b_layout const INIT_1 = {0, 1,1, 1,1 };
struct port_b_layout const INIT_2 = {3, 1,1, 1,0 };
struct port_b_layout const INIT_3 = {0, 0,0, 0,0 };
struct port_b_layout const FOR_SEND = {0,0,0, 0,0 };
// All outputs
struct port_b_layout const FOR_READ = {15,0,0, 0,0 };
// Data is an input
main() {
int x;
set_tris_b((int)FOR_SEND);
port_b = INIT_1;
delay_us(25);
// The constant
// structure is
// treated like
// a byte and
// is used to
// set the data
// direction
}
port_b = INIT_2;
port_b = INIT_3;
set_tris_b((int)FOR_READ);
port_b.rw=0;
port_b.cd=1;
port_b.enable=0;
x = port_b.data;
port_b.enable=0
// These constant structures delay_us(25);
// are used to set all fields
// on the port with a single
// command
// Here the individual
// fields are accessed
// independently.
How does the compiler determine TRUE and FALSE on
expressions?
When relational expressions are assigned to variables, the result is always 0 or 1.
For example:
bytevar = 5>0; //bytevar will be 1
bytevar = 0>5; //bytevar will be 0
The same is true when relational operators are used in expressions.
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