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.
What is the output of the code?
❌ D C
❌ E B
❌ D B
❌ KeyError B
✅ E KeyError
❌ None of the mentioned
Explanation:
Key-Value Indexing is used in the example above. D[1] = {‘A’ : {1 : “A”}, 2 : “B”}, D[1][2] = “B”, D[D[1][2]] = D[“B”] = “D” and D[“D”] = “E”. D[1] = {‘A’ : {1 : “A”}, 2 : “B”}, D[1][“A”] = {1 : “A”} and D[1][“A”][2] doesn’t exists, thus KeyError.
❌ D C
❌ E B
❌ D B
❌ KeyError B
✅ E KeyError
❌ None of the mentioned
Explanation:
Key-Value Indexing is used in the example above. D[1] = {‘A’ : {1 : “A”}, 2 : “B”}, D[1][2] = “B”, D[D[1][2]] = D[“B”] = “D” and D[“D”] = “E”. D[1] = {‘A’ : {1 : “A”}, 2 : “B”}, D[1][“A”] = {1 : “A”} and D[1][“A”][2] doesn’t exists, thus KeyError.