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C-Basics
quiz solution
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C Basics Tricks part 1
3
Operators priorities & Implicit Type Conversion
float g = 2 / 3.0 + 5 ;
float g = 2 / 3 + 5 ;
g =5.6
g =5
Signed Data type Range (+/-)
The sign of the result of a remainder operation, according to C99, is the same as
the dividend's one.
Let's see some examples (dividend / divisor):
And Two's complement
Asci with Octa number
10
Bit shifting is not multiplication. It can be used in
certain circumstances to have the same effect as a
multiplication by a power of two
addition(+) operator has higher precedence than
shift(<<) operator
1<<5<<4 == 1<<9 == 1*2^9 =512
Int i= 1,2,3 ๏ƒจ i= 3
Optimization will ignore the operation
Ans. 4 15
a==b==c (Multiple Comparison) evaluates in C
programming
(==) operates from left to right
Expression a==b==c is actually (a==b) ==c,
if(a==b==c)
Set Bit
X |= 1<<n ;
Reset specific Bit
X &= ~(1<<n) ;
Toggle Specific Bit
X ^= 1<<n ;
Implicit/
explicit
tricks
Prefix and postfix
13 12
evaluation takes place from left to right and
will be stopped if one of its components
evaluates to true(a non zero value).
Ans. 1 1 2 2
In case of GCC Compiler
Output will be 12,10. Output
may very from compiler to
compiler because order of
evaluation inside printf
The reason for undefined
behavior in PROGRAM 1 is,
the operator โ€˜+โ€™ doesnโ€™t have
standard defined order of
evaluation for its operands
1
2
3
4
5
6
7
8
9
10 11
12
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q1
4
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q2
5
๏ต Evaluate the following expressions
๏ต g = big / 2 + big * 4 / big - big + abc / 3 ; (abc = 2.5, big = 2, assume g to be a float)
๏ต on = ink * act / 2 + 3 / 2 * act + 2 + tig ; (ink = 4, act = 1, tig = 3.2, assume on to be an int)
๏ต s = qui * add / 4 - 6 / 2 + 2 / 3 * 6 / god ; (qui = 4, add = 2, god = 2, assume s to be an int)
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Think ๏Š
6
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Think ๏Š
7
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Think
8
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9
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q2
10
๏ต Evaluate the following expressions
๏ต g = big / 2 + big * 4 / big - big + abc / 3 ; (abc = 1.5, big = 3, assume g to be a float)
๏ต on = ink * act / 2 + 3 / 2 * act + 2 + tig ; (ink = 3, act = 2, tig = 3.2, assume on to be an int)
๏ต s = qui * add / 4 - 6 / 2 + 2 / 3 * 6 / god ; (qui = 2, add = 4, god = 3, assume s to be an int)
Two tricks
Int/int = int Float/int = float
Operation priorities
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q3
11Two tricks
Int/int = int Float/int = float
Operation priorities
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q3
12Two tricks
Int/int = int Float/int = float
Operation priorities
0 2 0.0 2.0
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Q4 no tricks
13
Compiler
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Q5/6/7/8 No tricks
14
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Q9 no tricks
15
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Q10
16
Two tricks
Int/int = int Float/int = float
Operation priorities
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Q10
17
Two tricks
Int/int = int Float/int = float
Operation priorities
7/22*5.14*3/5
0*5.14*3/5
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Q11
18
One trick
Signed int Range for 2 bytes
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Q11
19
One trick
Signed int Range for 2 bytes
First calculate the expression = 32768
If size of a signed data type is n bytes, it
ranges from -28n-1 to 28n-1-1
So, a short(usually 2 bytes) ranges from -32768
to 32767 and an unsigned short ranges from 0 to
65535
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Q12
20
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Think ๏Š
21
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Think ๏Š
22
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Q12
23
One trick
The sign of the result of a remainder operation, according to C99, is the same as the dividend's one.
Let's see some examples (dividend / divisor):
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q13
24
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q13
25
Two's complement
0111 1001 (+121)
-X
-X will be -121
Signed Data type Range (+/-)
One trick
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q14
26
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q14
27
X=0xFFFF FFFF
y=0xFFFF FFFF
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Q15 one trick
28
Int/int = int Float/int = float
Operation priorities
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Q15 one trick
29
Int/int = int Float/int = float
Operation priorities
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q16
30
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q16
31
Asci with Octa number
One trick
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q17
32
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Think ๏Š
33
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Think ๏Š
34
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q17
35
Bit shifting is not multiplication. It can be used in
certain circumstances to have the same effect as a
multiplication by a power of two
1<<5<<4 == 1<<9 == 1*2^9 =512
One trick
Explanation: The main logic behind the program is the
precedence and associativity of the operators. The
addition(+) operator has higher precedence than shift(<<)
operator. So, the expression boils down to 1 << (2 + 3) <<
4 which in turn reduces to (1 << 5) << 4 as the shift
operator has left-to-right associativity
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q18
36
Int i= 1,2,3 == i= 3
One trick
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q19
37
Optimization tricks
One trick
4 15
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q20
38
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Think ๏Š
39
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Think ๏Š
40
See the program, the values of a, b and c is 100 and you are thinking how
condition is false hereand why output is "False..."?
The expression is a==b==c which will evaluates like (a==b)==c now what will
be the result?
โ€ขThe result of (a==b) is 1 (i.e. true).
โ€ขAnd (1)==c will be 0 (i.e. false) because the value of c is 100 and 100 is
equal not to 1.
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q20
41
Output:
FALSE
a==b==c (Multiple Comparison) evaluates in C programming
(==) operates from left to right
Expression a==b==c is actually (a==b) ==c,
One trick
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q21
42
= 1
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Q22/23/24
43
reset
Set
toggle
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Implicit/explicit
tricks
44
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Type Conversion in C
45
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Implicit Type Conversion
46
Also known as โ€˜automatic type conversionโ€™.
โ€ขDone by the compiler on its own, without any external trigger from the user.
โ€ขGenerally takes place when in an expression more than one data type is present. In
such condition type conversion (type promotion) takes place to avoid lose of data.
โ€ขAll the data types of the variables are upgraded to the data type of the variable with
largest data type.
โ€ขIt is possible for implicit conversions to lose information, signs can be lost (when signed is
implicitly converted to unsigned), and overflow can occur (when long long is implicitly converted
to float).
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Explicit Type Conversion
47
This process is also called type casting and it is user defined. Here the user can type
cast the result to make it of a particular data type.
The syntax in C:
(type) expression
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Implicit/explicit tricks 48
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q30
49The reason for undefined behavior in PROGRAM 1 is, the operator โ€˜+โ€™
doesnโ€™t have standard defined order of evaluation for its operands
Explanation: See https://www.geeksforgeeks.org/sequence-points-in-c-set-1/ for explanation.
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q31
50
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q31
51
Prefix and postfix
One trick
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q32
52
Prefix and postfix
One trick
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q32
53
Explanation: In C, prefix and postfix operators
need l-valueto perform operation and return
r-value. The expression (++i)++ when executed
increments the value of variable i(i is a l-value) and
returns r-value. The compiler generates the error(l-
value required) when it tries to post-incremeny the
value of a r-value.
Prefix and postfix
One trick
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q33
54
In case of GCC Compiler Output will be 12,10. Output
may very from compiler to compiler because order of
evaluation inside printf
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q34
55
X = (j || (i++ && 1));
Note that precedence does not equate to order of
execution in general. In this case, we have the following
evaluation logic:
To evaluate = we need to evaluate its right-hand operand
To evaluate (j || stuff...) we first evaluate j
j is non-zero, so the result of (j || stuff...) is 1, and we do
not evaluate stuff due to the short-circuit behaviour of ||
Now we have determined that the right-hand operand of =
has evaluated to 1, so assign 1 to X.
Final result: X == 1, and i and j unchanged.
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q35
56
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q35
57
Answer: Option B
Solution:
In an expression involving || operator, evaluation takes
place from left to right and will be stopped if one of its
components evaluates to true(a non zero value).
So in the given expression m = i++ || j++ || k++.
It will be stop at j and assign the current value of j in m.
therefore m = 1 , i = 1, j = 2 and k = 2 (since k++ will not
encounter. so its value remain 2)
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q36
58
Answer: Option A
Solution:
-1 will be represented in binary form as:
1111 1111 1111 1111
Now -1<<4 means 1 is Shifted towards left by 4 positions,
hence it becomes:
1111 1111 1111 0000 in hexadecimal form - fff0.
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59
ENG. Keroles Shenouda
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