# Chapter 3: Control Flow S. M. Farhad. Statements and Blocks An expression becomes a statement when it is followed by a semicolon Braces { and } are used.

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Chapter 3: Control Flow S. M. Farhad

Statements and Blocks An expression becomes a statement when it is followed by a semicolon Braces { and } are used to group declarations and statements together into a compound statement or block A block is syntactically equivalent to a single statement Variables can be declared inside any block There is no semicolon after the right brace that ends a block

if-else The if-else statement is used to express decisions Formally the syntax is if (expression) statement 1 else statement 2 The else part is optional

if-else Contd. The statement 1 is executed if expression is true (having a non-zero value) The statement 2 is executed if expression is false (having a zero value) Since if simply tests the numeric value of an expression coding shortcuts are possible if (expression) and if (expression != 0) have the same results

if-else Contd. Dangling else problem z = c; if (n > 0) if (a > b) z = a; else z = b; As the else part of an if-else is optional, there is an ambiguity when an else is omitted from a nested if sequence This is resolved by associating the else with the closest previous else-less if It is good idea to use braces when there are nested ifs.

else-if if (expression) statement else if (expression) statement else if (expression) statement else statement The last else part handles the none of the above or default case (optional)

Binary Search If a particular value x occurs in the sorted array v The elements of v must be in increasing order The function returns the position if x occurs in v, and -1 if not

Binary Search Step 1 int binsearch(int x, int v[ ], int n){ int low, high, mid; low = 0; high = n -1; while(low <= high){ mid = (low + high)/2; if (x < v[mid]) high = mid – 1; else if(x > v[mid]) low = mid + 1; else return mid; } return -1; } 12345 12340 x = 4 low = 0, high = 5 – 1 = 4

Binary Search Step 1 int binsearch(int x, int v[ ], int n){ int low, high, mid; low = 0; high = n -1; while(low <= high){ mid = (low + high)/2; if (x < v[mid]) high = mid – 1; else if(x > v[mid]) low = mid + 1; else return mid; } return -1; } 12345 12340 low = 0, high = 4 mid = (0 + 4) / 2 = 2 x = 4 > v[2] = 3 low = 2 + 1 = 3 3

Binary Search Step 1 int binsearch(int x, int v[ ], int n){ int low, high, mid; low = 0; high = n -1; while(low <= high){ mid = (low + high)/2; if (x < v[mid]) high = mid – 1; else if(x > v[mid]) low = mid + 1; else return mid; } return -1; } 12345 12340 low = 3, high = 4 mid = (3 + 4) / 2 = 3 x = 4 = v[3] return 3 4

Switch The switch statement is a multi-way decision that tests whether an expression matches one of a number of constant integer values switch (expression){ case const-expr: statements default:statements } All case expressions must be different Default is executed if none of the cases match A default is optional

Switch Contd. The break statement causes an immediate exit from the switch Falling through cases is a mixed blessing Falling through one case to another is not robust Put a break after the last case even though it is unnecessary A number is n, if n is even then show the number is even if n is odd then show the number is odd

Loops-While and For for(expr1; expr3; expr3) statement is equivalent to expr1; while (expr2) { statement expr3; }

Loops-While and For Although equivalent the syntax of while is while (expr) stmt

What We Practice If-else if (expression){ statement 1 } else{ statement 2 } Similarly else-if

What We Practice Contd. Loops for(expr1; expr3; expr3) { statement } while (expr) { statement }

Loops-While and For Contd. While is most natural when there is no initialization or re-initialization For is preferable there is a simple initialization and increment Since it keeps the loop control stmts close together and visible at the top of the loop The index and limit of a C for loop can be altered from within the loop The index variable retains its value when the loop terminates any reason The for loop is not restricted to arithmetic progressions It is bad style to force unrelated computations into the initialization and increment portion

int atoi(char s[ ]) convert s to integer int atoi(char s[ ]){ int i, n, sign; for(i = 0; isspace(s[i]); i++) ; /* skip white space */ sign = (s[i] == -) ? -1 : 1; if(s[i] == + || s[i] == -) i++; /* skip sign */ for(n = 0; isdigit(s[i]); i++) n = 10 * n + (s[i] – 0); return sign * n; } -345 01234 \0 5 i n = 10*0 + (51 – 48) = 3

int atoi(char s[ ]) convert s to integer int atoi(char s[ ]){ int I, n, sign; for(i = 0; isspace(s[i]); i++) ; /* skip white space */ sign = (s[i] == -) ? -1 : 1; if(s[i] == + || s[i] == -) i++; /* skip sign */ for(n = 0; isdigit(s[i]); i++) n = 10 * n + (s[i] – 0); return sign * n; } -345 01234 \0 5 i n = 10*3 + (52 – 48) = 34

int atoi(char s[ ]) convert s to integer int atoi(char s[ ]){ int I, n, sign; for(i = 0; isspace(s[i]); i++) ; /* skip white space */ sign = (s[i] == -) ? -1 : 1; if(s[i] == + || s[i] == -) i++; /* skip sign */ for(n = 0; isdigit(s[i]); i++) n = 10 * n + (s[i] – 0); return sign * n; } -345 01234 \0 5 i n = 10*34 + (53 – 48) = 345

Nested Loop: Bubble Sort void bubble(int s[ ], n){ int i, j, temp; for(i = 1; i < n; i++) for(j = n - 1; j >= i; j--) if(s[j-1] > s[j]){ temp = s[j-1]; s[j-1] = s[j]; s[j] = temp; } 82541 12340

Nested Loop: Bubble Sort Contd. void bubble(int s[ ], n){ int i, j, temp; for(i = 1; i < n; i++) for(j = n - 1; j >= i; j--) if(s[j-1] > s[j]){ temp = s[j-1]; s[j-1] = s[j]; s[j] = temp; } 82514 12340

Nested Loop: Bubble Sort Contd. void bubble(int s[ ], n){ int i, j, temp; for(i = 1; i < n; i++) for(j = n - 1; j >= i; j--) if(s[j-1] > s[j]){ temp = s[j-1]; s[j-1] = s[j]; s[j] = temp; } 82154 12340

Nested Loop: Bubble Sort Contd. void bubble(int s[ ], n){ int i, j, temp; for(i = 1; i < n; i++) for(j = n - 1; j >= i; j--) if(s[j-1] > s[j]){ temp = s[j-1]; s[j-1] = s[j]; s[j] = temp; } 81254 12340

Nested Loop: Bubble Sort Contd. void bubble(int s[ ], n){ int i, j, temp; for(i = 1; i < n; i++) for(j = n - 1; j >= i; j--) if(s[j-1] > s[j]){ temp = s[j-1]; s[j-1] = s[j]; s[j] = temp; } 18254 12340

Try Output void pyramid(int n){ int i, j, k; for(i = 0; i < n; i++){ for(j = 0; j < (n - i); j++ ){ printf(" "); // space } for(k = 0, j = i + 1; k <= i; j++, k++){ printf("%d ", j % 10);//1st side of a row } for(k = 0, j -= 2; k < i; j--, k++){ printf("%d ", j % 10);//2nd side of a row } printf("\n"); } } RunRun

Loops--Do-while The syntax do stmt while (exp); Tests the termination condition at the bottom after making each pass through the loop body The body is always executed at least once The sequence of execution

Loops--Do-while Contd. Do-while is much less used than while and for From time to time it is valuable Itoa function

Itoa: convert n to string s void itoa (int n, char s[ ]){ if ((sign = n) < 0) n = -n;/* make it positive */ i = 0; do{ s [i++] = n % 10; + 0; } while ((n /= 10) > 0); if (sign < 0) s [i++] = -; s [i] = \0; reverse (s); } Re write the do-while loop using while loop so that Integrity is same Menu Option Correction

Break and Continue It is sometimes convenient to be able to exit from a loop other than by testing at the top or bottom break statement provides an early exit from for, while, and do, just as from switch A break causes the innermost enclosing loop or switch to be exited immediately The function trim removes the trailing blanks, tabs, and new lines from the end of a string

Trim: remove trailing blanks, tabs, newlines int trim (char s [ ]){ int n; for (n = strlen (s) – 1; n >= 0; n--) if (s [n] != && s [n] = \t && s [n] != \n) break; s [n + 1] = \0; return n; } Returns the length of The string s Breaks the loop when any char other than white spaces When the entire string has been scanned

Continue Statement It is related to break statement but less often used It causes the next iteration of the enclosing for, while, or do loop to begin In the case of while and do, this means that the test part is executed immediately In the for, control passes to the increment step The continue statement applies only to loops, not to switch

An Example with continue The following fragment processes only the non-negative elements in the array a; negative values are skipped for (i = 0; i < n; i++){ if (a[i] < 0) continue;/* skip negative elements */..... /* do positive elements */ }

What is the output? for (i = 1; i < 5; i++){ for (j = 1; i < 5; j++){ if (i == j) break; printf(%d %d, i, j); } printf (\n) }

What is the output? With Answer for (i = 1; i < 5; i++){ for (j = 1; i < 5; j++){ if (i == j) break; printf(%d %d, i, j); } printf (\n) } 1 2 1 2 2 3 1 3 2 3 3 4 1 4 2 4 3 4 4

What is the output? for (i = 1, x= 0; i <= 4; i++){ for (j = 1; i <= 3; j++){ if (i == j) continue; printf (i = %d j = %d,, i, j); x = x + i +j; } printf (\nx = %d, x); } printf (\nx = %d, x);

What is the output? for (i = 1, x= 0; i <= 4; i++){ for (j = 1; i <= 3; j++){ if (i == j) continue; printf (i = %d j = %d,, i, j); x = x + i +j; } printf (\nx = %d, x); } printf (\nx = %d, x); i=1 j=2, i=1 j=3 x = 7 i=2 j=1, i=2 j=3 x = 15 i=3 j=1, i=3 j=2 x = 24...

Goto and Labels C provides infinitely-abusable goto statement and labels to branch to Formally, the goto is never necessary In practice, it is always easy to write code without it We will not allow to use goto statement to used in our lab assignments

Goto and Labels Contd. Nevertheless, there are a few situations where gotos may find a place The most common is to abandon processing in some deeply nested structure Breaking out of two or more loops at once Code that relies on got statements is generally harder to understand and to maintain than cod e without gotos

Goto and Labels Contd. for (.. ) for (.. ) { … if (disaster) goto error; }... error: /* clean up the mess */

Goto and Labels Contd. As another example, consider the problem of determining whether two arrays a and b have an element in common for (i = 0, i < n; i++) for (j = 0, j < m; j++) if (a [i] == b [j]) goto found; found:...

The End of Chapter 3 Any Question?

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