Expressions So far, we have seen fairly simple expressions. For example, x = 4; z = z - y; if(x%4 == 0) However, expressions can be much more complex.

Expressions – Complex Here is an example of a complex expression being used in a for loop. From page 41 of The C Programming Language, 2nd. Ed. by Kernighan and Ritchie, for(i=0; i<lim-1 && (c=getchar()) != ’\n’ && c != EOF; ++i) s[i] = c; We can see that the expressions for initializing and incrementing the counter are simple. The complexity occurs in the middle expression for testing whether to continue the loop.

Expressions Sometimes a complex expression is less cumbersome than the equivalent using only simple expressions. if (a && b && c && d && e) x = 5; does the same thing as if (a) if (b) if (c) if (d) if (e) x = 5;

Expressions – Examples To use more complex expressions, we often must either have knowledge of the order of precedence of the operators control the order of precedence using parentheses Recall the example from Kernighan and Ritchie for determining if a year is a leap year: if( year%4 == 0 && year%100 != 0 || year%400 == 0 )

Increment/Decrement, Part 2 We saw how to use the increment (++) and decrement (- -) operators to increase or decrease a variable’s value by 1, e.g., sum++. We can place the double plus or minus signs before the variable (prefix) or after the variable (postfix). Depending on where a variable is being used when incremented/decremented, this can make a difference.

Increment/Decrement, Part 2 The increment and decrement operators are unary operators, so they require one argument. We can use them to modify a variable’s value when that variable is part of a larger expression. Examples: x++; /* makes no difference */ ++x; /* makes no difference */ sum = sum + y++; /* y is added to sum, then y is incremented */ sum = sum + ++y; /* y is incremented, then the new value of y is added to sum */

Increment/Decrement, Part 2 Here we see equivalent methods for using a variable before incrementing it: x = 3 + y++;  x = 3 + y; y++; Here we see equivalent methods for incrementing a variable before using it: x = y;  y++; x = 3 + y;

Increment/Decrement, Part 2 This code int x = 3; printf("line 1: x is %d\n", x); printf("line 2: x is %d\n", x++); /* increment x after using */ printf("line 3: x is %d\n", x); printf("line 4: x is %d\n", ++x); /* increment x before using */ printf("line 5: x is %d\n", x); produces line 1: x is 3 line 2: x is 3 line 3: x is 4 line 4: x is 5 line 5: x is 5

Incrementing in for Loops int i; for(i = 0; i <= 5; i++) /* increment as postfix */ printf("%d ", i); for(i = 0; i <= 5; ++i) /* increment as postfix */ printf("%d ", i); for(i = 5; i >= 0; i--) /* increment as postfix */ printf("%d ", i); for(i = 5; i >= 0; --i) /* increment as postfix */ printf("%d ", i); produces

Incrementing in for Loops Why doesn’t it matter in a for loop if we use the postfix or prefix versions? Remember that for a for loop we can create an equivalent while loop. int i, x = 0; for(i = 0; i < 5; i++) x = x + i; is equivalent to int i, x = 0; i = 0; while(i < 5) { x = x + i; i++; }

Functions in Expressions When we call a function that returns a value, that value is used in the expression. y = 4; x = x + doubleNum(y); /* assume doubleNum returns 8 */ This is equivalent to y = 4; x = x + 8; This means we can use functions in expressions just as we would use other values.

Functions in Expressions Here we use a function in the typical way. #include int squareNum(int num) { return num*num; } int main(void) { int x = 1; int sq = squareNum(x); while( sq <=100 ) { printf("The square of %2d is %3d\n", x, sq); x++; sq = squareNum(x); }

Functions in Expressions This program has the same functionality but includes the function in the test of the while loop. #include int squareNum(int num) { return num*num; } int main(void) { int x = 1, sq; while( (sq = squareNum(x)) <= 100 ) { printf("The square of %2d is %3d\n", x, sq); x++; }

Functions in Expressions #include int doubleNum(int num) { return 2*num; } int main(void) { int x = 8, output; output = doubleNum( doubleNum(x) ); printf("output is %d\n", output); /* output is 32 */ } To evaluate, we must work our way inside much the way we do when looking at mathematical expressions with many parentheses, e.g., (((x − 3) − 2x 2 ) + 8)