Mixing
Stack could not be converted to int
-1
I have below code
public function account(Stack $volume) {
//echo $volume->value(); //This line prints 300
//echo $this -> balance(); //This line prints 400
//echo gettype($volume -> value()); //int
//echo gettype($this->balance()); //object
echo $this -> balance() + $volume -> value(); // This line prints "Notice: Object of class Stack could not be converted to int"
}
Why it is happening like this ?
php oop
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
|
show 6 more comments
-1
I have below code
public function account(Stack $volume) {
//echo $volume->value(); //This line prints 300
//echo $this -> balance(); //This line prints 400
//echo gettype($volume -> value()); //int
//echo gettype($this->balance()); //object
echo $this -> balance() + $volume -> value(); // This line prints "Notice: Object of class Stack could not be converted to int"
}
Why it is happening like this ?
php oop
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
1
Just add()to clarify your logic.
– u_mulder
Nov 21 '18 at 7:52
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:($volume->value())
– Snake14
Nov 21 '18 at 8:01
3
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
2
I see the results of thegettype()and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?
– Rei
Nov 21 '18 at 9:55
|
show 6 more comments
-1
-1
-1
1
I have below code
public function account(Stack $volume) {
//echo $volume->value(); //This line prints 300
//echo $this -> balance(); //This line prints 400
//echo gettype($volume -> value()); //int
//echo gettype($this->balance()); //object
echo $this -> balance() + $volume -> value(); // This line prints "Notice: Object of class Stack could not be converted to int"
}
Why it is happening like this ?
php oop
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
I have below code
public function account(Stack $volume) {
//echo $volume->value(); //This line prints 300
//echo $this -> balance(); //This line prints 400
//echo gettype($volume -> value()); //int
//echo gettype($this->balance()); //object
echo $this -> balance() + $volume -> value(); // This line prints "Notice: Object of class Stack could not be converted to int"
}
Why it is happening like this ?
php oop
php oop
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
edited Nov 21 '18 at 9:35
abu abu
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
asked Nov 21 '18 at 7:45
abu abuabu abu
1,14221444
1,14221444
1
Just add()to clarify your logic.
– u_mulder
Nov 21 '18 at 7:52
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:($volume->value())
– Snake14
Nov 21 '18 at 8:01
3
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
2
I see the results of thegettype()and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?
– Rei
Nov 21 '18 at 9:55
|
show 6 more comments
1
Just add()to clarify your logic.
– u_mulder
Nov 21 '18 at 7:52
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:($volume->value())
– Snake14
Nov 21 '18 at 8:01
3
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
2
I see the results of thegettype()and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?
– Rei
Nov 21 '18 at 9:55
1
1
Just add
() to clarify your logic.– u_mulder
Nov 21 '18 at 7:52
Just add
() to clarify your logic.– u_mulder
Nov 21 '18 at 7:52
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:
($volume->value())– Snake14
Nov 21 '18 at 8:01
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:
($volume->value())– Snake14
Nov 21 '18 at 8:01
3
3
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
2
2
I see the results of the
gettype() and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?– Rei
Nov 21 '18 at 9:55
I see the results of the
gettype() and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?– Rei
Nov 21 '18 at 9:55
|
show 6 more comments
1 Answer
1
active
oldest
votes
3
Taking the clues from these lines of your code:
//echo $this -> balance(); //This line prints 400
...
//echo gettype($this->balance()); //object
That means the object returned by $this->balance() can be converted to a string with a numeric value, which is 400 in your code.
To convert that to an int, you can use strval() followed by intval() like this:
echo intval(strval($this -> balance())) + $volume -> value();
Because the + operator can also work with numeric values in strings, just strval() will also work:
echo strval($this -> balance()) + $volume -> value();
The choice is yours.
answered Nov 21 '18 at 10:13
ReiRei
5,551723
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?(int) (string) $Account->balance() == 300
– abu abu
Nov 21 '18 at 10:26
2
@abuabu The==operator compares the left and right operands for equality. The left operand does the same thing asintval(strval($Account->balance())).
– Rei
Nov 21 '18 at 10:48
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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oldest
votes
3
Taking the clues from these lines of your code:
//echo $this -> balance(); //This line prints 400
...
//echo gettype($this->balance()); //object
That means the object returned by $this->balance() can be converted to a string with a numeric value, which is 400 in your code.
To convert that to an int, you can use strval() followed by intval() like this:
echo intval(strval($this -> balance())) + $volume -> value();
Because the + operator can also work with numeric values in strings, just strval() will also work:
echo strval($this -> balance()) + $volume -> value();
The choice is yours.
answered Nov 21 '18 at 10:13
ReiRei
5,551723
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?(int) (string) $Account->balance() == 300
– abu abu
Nov 21 '18 at 10:26
2
@abuabu The==operator compares the left and right operands for equality. The left operand does the same thing asintval(strval($Account->balance())).
– Rei
Nov 21 '18 at 10:48
add a comment |
3
Taking the clues from these lines of your code:
//echo $this -> balance(); //This line prints 400
...
//echo gettype($this->balance()); //object
That means the object returned by $this->balance() can be converted to a string with a numeric value, which is 400 in your code.
To convert that to an int, you can use strval() followed by intval() like this:
echo intval(strval($this -> balance())) + $volume -> value();
Because the + operator can also work with numeric values in strings, just strval() will also work:
echo strval($this -> balance()) + $volume -> value();
The choice is yours.
answered Nov 21 '18 at 10:13
ReiRei
5,551723
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?(int) (string) $Account->balance() == 300
– abu abu
Nov 21 '18 at 10:26
2
@abuabu The==operator compares the left and right operands for equality. The left operand does the same thing asintval(strval($Account->balance())).
– Rei
Nov 21 '18 at 10:48
add a comment |
3
3
3
Taking the clues from these lines of your code:
//echo $this -> balance(); //This line prints 400
...
//echo gettype($this->balance()); //object
That means the object returned by $this->balance() can be converted to a string with a numeric value, which is 400 in your code.
To convert that to an int, you can use strval() followed by intval() like this:
echo intval(strval($this -> balance())) + $volume -> value();
Because the + operator can also work with numeric values in strings, just strval() will also work:
echo strval($this -> balance()) + $volume -> value();
The choice is yours.
answered Nov 21 '18 at 10:13
ReiRei
5,551723
Taking the clues from these lines of your code:
//echo $this -> balance(); //This line prints 400
...
//echo gettype($this->balance()); //object
That means the object returned by $this->balance() can be converted to a string with a numeric value, which is 400 in your code.
To convert that to an int, you can use strval() followed by intval() like this:
echo intval(strval($this -> balance())) + $volume -> value();
Because the + operator can also work with numeric values in strings, just strval() will also work:
echo strval($this -> balance()) + $volume -> value();
The choice is yours.
answered Nov 21 '18 at 10:13
ReiRei
5,551723
answered Nov 21 '18 at 10:13
ReiRei
5,551723
answered Nov 21 '18 at 10:13
ReiRei
5,551723
answered Nov 21 '18 at 10:13
ReiRei
5,551723
5,551723
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?(int) (string) $Account->balance() == 300
– abu abu
Nov 21 '18 at 10:26
2
@abuabu The==operator compares the left and right operands for equality. The left operand does the same thing asintval(strval($Account->balance())).
– Rei
Nov 21 '18 at 10:48
add a comment |
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?(int) (string) $Account->balance() == 300
– abu abu
Nov 21 '18 at 10:26
2
@abuabu The==operator compares the left and right operands for equality. The left operand does the same thing asintval(strval($Account->balance())).
– Rei
Nov 21 '18 at 10:48
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
Many Many Thanks @Rei. Your solution is working like a Magic.
– abu abu
Nov 21 '18 at 10:21
1
1
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
@abuabu You're welcome.
– Rei
Nov 21 '18 at 10:22
can you say what does it mean ?
(int) (string) $Account->balance() == 300– abu abu
Nov 21 '18 at 10:26
can you say what does it mean ?
(int) (string) $Account->balance() == 300– abu abu
Nov 21 '18 at 10:26
2
2
@abuabu The
== operator compares the left and right operands for equality. The left operand does the same thing as intval(strval($Account->balance())).– Rei
Nov 21 '18 at 10:48
@abuabu The
== operator compares the left and right operands for equality. The left operand does the same thing as intval(strval($Account->balance())).– Rei
Nov 21 '18 at 10:48
add a comment |
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1
Just add
()to clarify your logic.– u_mulder
Nov 21 '18 at 7:52
Please provide the object as well?
– Tausif Anwar
Nov 21 '18 at 7:54
It's probably trying to add $volume to the balance instead of the value from the $volume. Surrounding the accessor with parenthesis should fix your issue. I'd also recommend getting rid of the extra spaces in between, like the following:
($volume->value())– Snake14
Nov 21 '18 at 8:01
3
Cannot reproduce the error. The problem must be somewhere else. See 3v4l.org/JWjHn
– Rei
Nov 21 '18 at 9:06
2
I see the results of the
gettype()and that means you're trying to add an object to an int. That is an error. Are you expecting that it should not be an error?– Rei
Nov 21 '18 at 9:55