Unable to update the value of variable [duplicate] - c++

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?


Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]


Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.


Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).


Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors


Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0


I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

Related

it isn't generating random steps [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?

Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).

Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors

Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0

I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

C++ Function is Producing the Same Values in a Vector Everytime it is Called [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?

Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).

Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors

Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0

I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

Why won't the pi variable update with the loop? [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?

Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).

Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors

Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0

I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

Why Does The Math Formula Not Work In C++? [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?

Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).

Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors

Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0

I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

Matrix is zero after assignment. C++Armadillo [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?

Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).

Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors

Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0

I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.