The values of the following variables are a, b, c and number of memory leaks in the program below is:void ModifyVariables(int a, int &b, int *c){ a = b; b+=a;; c = new int(b); (*c)++;}int main(){ int a=0,b=1, *c; c = new int(2); ModifyVariables(a,b,c); cout<<a<<b<<*c; delete c;}
Question
The values of the following variables are a, b, c and number of memory leaks in the program below is:void ModifyVariables(int a, int &b, int *c){ a = b; b+=a;; c = new int(b); (*c)++;}int main(){ int a=0,b=1, *c; c = new int(2); ModifyVariables(a,b,c); cout<<a<<b<<*c; delete c;}
Solution
The number of memory leaks in the program is 1.
Here's the step by step explanation:
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In the main function, memory is allocated for the integer pointer 'c' using the 'new' operator. This memory is not deallocated within the main function, which is fine because it is deallocated later.
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The function ModifyVariables is called with 'a', 'b', and 'c' as arguments. Inside this function, 'c' is assigned a new memory location with 'new' operator. This is where the memory leak occurs. The original memory location that 'c' was pointing to is lost without being deallocated.
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At the end of the main function, 'delete c' is called. This deallocates the memory assigned to 'c' in the ModifyVariables function, but the memory initially allocated in the main function is never deallocated, hence causing a memory leak.
So, the memory initially allocated to 'c' in the main function is never deallocated, causing a memory leak.
Similar Questions
void ModifyVariables(int a, int &b, int *c){ a = b; b+=a;; c = new int(b); (*c)++;}int main(){ int a=0,b=1, *c; c = new int(2); ModifyVariables(a,b,c); cout<<a<<b<<*c; delete c;}
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