Edhesive 6.8 lesson practice answers
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Answer:
1.) 25 ; 15 ; 15
2.) 50 ; 15 ; 50
Explanation:
In the first function written :
The variable val was initially decaled or assigned a value of 25 and that was what was printed first.
However, after the example function was written, the val variable was finally assiagned a value of 15 within the function. However, it was also declared that the global variable takes uonthe val value. Hence, the val variable initially assigned a value, of 25 changes to 15 globally.
For the second code :
From the top:
Val was assigned a value of 50 ;
Hence,
print(val) gives an output of 50
Within the function definition which prints the value of val that is assigned a value of 25 within the function.
Since tbe global variable isnt reset.
Printing Val again outputs 50;since ito is outside the function.
The code output description can be defined as follows:
Code Explanation:
Code:
val = 25 #defining a variable val that holds an integer value
def example(): #defining a method example
global val #using keyword global to define variable val
val = 15#defining an integer variable that hold integer value
print (val)#print global variable value
print (val) #print val variable value
example()#calling example method
print (val)#print global variable value
print("----------------------------------------------------")
val = 50 #defining a variable val that holds an integer value
def example(): #defining a method example
val = 15 #define variable val that holds integer value
print(val)#print val variable value
print (val) #print val variable value
example() #calling method example
print (val)#print val value
Output:
Please find the attached file.
Learn more:
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