(制御文: 条件、枝分れ、繰返し)
http://www.sw.it.aoyama.ac.jp/2017/CP1/lecture3.html
© 2005-17 Martin J. Dürst 青山学院大学
| 02B1 | 02C1 | 02C2 | |
| 100 points | 99 | 84 | 63 |
| 60 points | - | 15 | 35 |
| not submitted | - | - | 1 |
Examples of how to check for odd numbers:
a-((a>>1)<<1)==0
(a>>1)<<1 == 1
One more example: a & 1 == 1
Intent: (a&1) == 1
Actual interpretation: a & (1==1) (see p.
388)
Accidental luck: The result of 1==1 (true) by chance was 1
(this is not guaranteed)
Conclusion 1: Please be careful with the priority of shift and bitwise operations
Conclusion 2: a & 1 is enough
(for even numbers, the result is 0 (false), for odd numbers, the result is 1
(true))
int a, b, c, d, e;buffer is better than
buf)leap_year is better than uruudoshi)if statement:if ( condition )
statement
Grammar consists of fixed
parts (if、(、and ) above;
terminals) and replacable parts (nonterminals)
if statement:Evaluate the condition. If the result is true, then
execute the statement. Otherwise, do not execute the
statement.
a = 20;
if (a < 10)
a = 15;
The if statement is syntactically correct, but semantically
wrong.
Syntactic errors are caught by the compiler. Most semantic errors are not caught by compilers.
式; (expression statement)if statement,switch statement, for
statement, while statementreturn statement, break statement{})This will lead to errors:
if (score >= 60)
printf("You got 60 or more points.\n");
printf("You passed the exam.\n");
This is much clearer:
if (score >= 60) {
printf("You got 60 or more points.\n");
printf("You passed the exam.\n");
}
if and elseif (score >= 60) {
printf("You got more than 60 points.\n");
} else {
printf("See you again next year.\n");
}
if - else if -
elseelseif or elsif or
elif if - else if - else
if - ... - else as a single structureelse if/else are indented at the
same level as the original if:if (...) {
...
}
else if (...) {
...
}
else {
...
}
if -
else if - elseif (...) {
...
}
else if (...) {
...
}
else if (...) {
...
}
else {
...
}
if: Ternary Operator(also called conditional operator)
Frequent example:
if (x > y)
max = x;
else
max = y;
Replacement:
max = (x > y) ? x : y;
if: Conditional ExpressionsUsing && or ||, it is possible to create
replacements for if statements:
if (score >= 60) {
grade = calculate(score);
}
Replacement:
score>=60 && (grade=calculate(score));
(Caution: not recommended in C!)
Example 1: gcc 03B1.c && ./a
Example 2: gcc -o 03B1 03B1.c && ./03B1
How it works:
It is also possible to connect two different executions in this way.
The result of execution is the value returned from main (0 if
successful).
Outside a program (i.e. in the (cygwin) shell), 0 is
true and other values are false.
Inside a C program, 0 is false and other
values are true.
gcc 02B1.c && ./a を推奨)switch Statementswitch ( expression ) {
case literal1 :
処理1 ;
break;
case literal2:
case literal3:
処理2 ;
...
break;
default :
処理3 ;
break;
}
switch 文の注意点Without a break statement, execution continues past case
labels.
int a = 5, b = 1;
switch (a) {
case 4:
b = 3;
break;
case 5:
b = 7;
case 6:
b *= 3;
break;
}
After execution, b is 21
If break is purposely omitted, use a comment
breakcase part of a switch statementif と switch の使い分け主な目安:
ifswitch を検討switch を検討for Statementfor ( 初期化; 継続条件; 次の一歩 ) {
繰り返す処理;
}
while Statementwhile ( 継続条件 ) {
繰り返す処理;
}
for と while の使い分け主な目安:
forforwhiledo ... while Statement文法:
do {
繰り返す処理;
} while ( 継続条件 );
(Russian peasant multiplication)
a0, b0: original multiplier and multiplicand (input)
sf: Final total (output)
Goal: sf = a0 · b0
Loop invariant: ai · bi + si = a0 · b0
Induction base:
For i = 0, through initialization, s0 = 0, and
therefore
a0 · b0 + s0
= a0 · b0 + 0 =
a0 · b0
What we have to prove: The invariant holds at all i (0 ≦ i ≦ f)
Inductive step:
Assume the invariant holds at the end of loop pass k 、(and therefore at the start of loop pass k+1)
Assumption: ak · bk + sk = a0 · b0
Ifak is even: a0 · b0 = ak · bk + sk = ak/2 · bk·2 + sk = ak+1 · bk+1 + sk+1 = a0 · b0 (/ is integer division)
If ak is odd: a0 · b0 = ak · bk + sk = ak/2 · bk·2 + bk + sk = ak+1 · bk+1 + sk+1 = a0 · b0
Termination condition: ai = 0
Because ai+1 = ai / 2, ai is reduced to half in each loop pass, and will reach 0 eventually.
Result check:
When ai = 0, a0 · b0 = ai · bi + si = 0 · bi + si = 0 + si = si
Therefore, when ai = 0, si is the correct result.
switch を使用)Example display: