What will be the output of the code?
❌ 113
❌ 14
❌ 12
✅ Error
❌ None of the mentioned
Explanation:
As you can see the variable ‘b’ is of type integer and the variable ‘a’ is of type string. Also as Python is a strongly typed language we cannot simply concatenate an integer with a string. We have to first convert the integer variable to the type string to concatenate it with a string variable. So, trying to concatenate an integer variable to a string variable, an exception of type “TypeError” is occurred.
❌ 113
❌ 14
❌ 12
✅ Error
❌ None of the mentioned
Explanation:
As you can see the variable ‘b’ is of type integer and the variable ‘a’ is of type string. Also as Python is a strongly typed language we cannot simply concatenate an integer with a string. We have to first convert the integer variable to the type string to concatenate it with a string variable. So, trying to concatenate an integer variable to a string variable, an exception of type “TypeError” is occurred.
What will be output for the code?
❌ A. Error, the syntax of the invoking method is wrong
❌ B. The program runs fine but nothing is printed
❌ C. 1 0
✅ D. 1 2
Explanation:
In the above piece of code, the invoking method has been properly implemented and hence x=1 and y=2.
❌ A. Error, the syntax of the invoking method is wrong
❌ B. The program runs fine but nothing is printed
❌ C. 1 0
✅ D. 1 2
Explanation:
In the above piece of code, the invoking method has been properly implemented and hence x=1 and y=2.
What will be output for the code?
❌ 1
❌ 2
✅ 3
❌ 4
❌ Code does not compile
❌ Error will occur
Explanation:
In Python the precedence order is first NOT then AND and in last OR. So the if condition and second elif condition evaluates to False while third elif condition is evaluated to be True resulting in 3 as output.
❌ 1
❌ 2
✅ 3
❌ 4
❌ Code does not compile
❌ Error will occur
Explanation:
In Python the precedence order is first NOT then AND and in last OR. So the if condition and second elif condition evaluates to False while third elif condition is evaluated to be True resulting in 3 as output.
What will be output for the code?
❌ 2 GFGNO
❌ 2.33 GFGNOG
❌ 2.33 GFGNONoneGTrue
✅ 2.33 GFGNO
❌ Error
❌ None of mentioned
Explanation:
val1 will only have integer and floating values val1 = 1 + 1.33 + 0 = 2.33 and val2 will have string values val2 =’GFG’ + ‘NO’ = ‘GFGNO’. String ‘G’ will not be part of val2 as the for loop will break at None, thus ‘G’ will not be added to val2.
❌ 2 GFGNO
❌ 2.33 GFGNOG
❌ 2.33 GFGNONoneGTrue
✅ 2.33 GFGNO
❌ Error
❌ None of mentioned
Explanation:
val1 will only have integer and floating values val1 = 1 + 1.33 + 0 = 2.33 and val2 will have string values val2 =’GFG’ + ‘NO’ = ‘GFGNO’. String ‘G’ will not be part of val2 as the for loop will break at None, thus ‘G’ will not be added to val2.
What will be output for the code?
❌ 1 1 1
❌ 2 2 2
❌ 3 3 3
❌ 1 2 3
✅ 3 2 3
❌ Error
❌ None of mentioned
Explanation:
In Python, class variables are internally handled as dictionaries. If a variable name is not found in the dictionary of the current class, the class hierarchy (i.e., its parent classes) are searched until the referenced variable name is found, if the variable is not found error is being thrown.
The output 3 2 3 may be surprising. Instead of 3 3 3, here B.val reflects 2 instead of 3 since it is overridden earlier.
❌ 1 1 1
❌ 2 2 2
❌ 3 3 3
❌ 1 2 3
✅ 3 2 3
❌ Error
❌ None of mentioned
Explanation:
In Python, class variables are internally handled as dictionaries. If a variable name is not found in the dictionary of the current class, the class hierarchy (i.e., its parent classes) are searched until the referenced variable name is found, if the variable is not found error is being thrown.
The output 3 2 3 may be surprising. Instead of 3 3 3, here B.val reflects 2 instead of 3 since it is overridden earlier.
What will be output for the code?
❌ {0: 0, 1: 0, 2: 0}
✅ {0: 1, 1: 1, 2: 1}
❌ {0: 0, 1: 0, 2: 0, 0: 1, 1: 1, 2: 1}
❌ TypeError: Immutable object
❌ Compilation error
❌ None of the mentioned
Explanation:
1st loop will give 3 values to i 0, 1 and 2. In the empty dictionary, valued are added and overwritten in j loop, for eg. D[0] = [0] becomes D[0] = 1, due to overwriting.
❌ {0: 0, 1: 0, 2: 0}
✅ {0: 1, 1: 1, 2: 1}
❌ {0: 0, 1: 0, 2: 0, 0: 1, 1: 1, 2: 1}
❌ TypeError: Immutable object
❌ Compilation error
❌ None of the mentioned
Explanation:
1st loop will give 3 values to i 0, 1 and 2. In the empty dictionary, valued are added and overwritten in j loop, for eg. D[0] = [0] becomes D[0] = 1, due to overwriting.
What is the output of the code?
❌ TypeError
❌ Code doesn't compile
❌ (1, 5, 3, 4)
❌ 1 TypeError
✅ 6 TypeError
❌ None of the mentioned
Explanation:
T1 is an integer while T2 is tuple. Thus T1 will become 1 + 5 = 6. But an integer and tuple cannot be added, it will throw TypeError.
❌ TypeError
❌ Code doesn't compile
❌ (1, 5, 3, 4)
❌ 1 TypeError
✅ 6 TypeError
❌ None of the mentioned
Explanation:
T1 is an integer while T2 is tuple. Thus T1 will become 1 + 5 = 6. But an integer and tuple cannot be added, it will throw TypeError.