The three best practices for Karim to build a professional network are:
The Best Practises?Listen to a contact's story with attention and empathy: People appreciate when they feel heard and understood. By actively listening to others, Karim can demonstrate his interest in them and their work, which can lead to a stronger relationship.
Provide a solution to a contact's immediate problem: If Karim can offer a helpful solution to a contact's problem, he is likely to be remembered as someone who is resourceful and knowledgeable, which can benefit his career in the long run.
Share a story about his plans and aspirations: By sharing his own aspirations and goals, Karim can create a connection with others who may have similar interests or career paths. This can help him expand his network and potentially lead to future opportunities.
Walking around meeting as many contacts as possible and leaving quickly with a brief nod after meeting a contact may not leave a lasting impression or lead to a strong relationship. It is better to focus on quality over quantity when building a professional network.
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Please Help. This is a C++ Program
Perform a walkthrough of the code below by filling in the value of the variable that is updated at each
step. Use the notation ->var to indicate that the value of the variable is a pointer to var. For the last two
lines in the table, show the values of the variables that are output in the Output column.
The table that shows the values the walkthrough of the code below by filling in the value of the variable that is updated at each step is given in the image attached.
What is the C++ Program?In step 1 of the table, the variable int1 is initialized to 26 and int2 is initialized to 45. In step 2, the values of int1 and int2 stay unaltered.
In step 3, the value pointed to by int1Ptr is changed to 89 is using the * administrator. In step 4, the esteem pointed to by int2Ptr is changed to 62 utilizing the * administrator.
In step 5, int1Ptr is assigned to int2Ptr, linking both pointers to the same memory location (&int2). Int1 and int2 values remain unchanged. In step 6, int1Ptr points to a memory location with a value of 80 after using the * operator.
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See table below
int1 int2 *int1Ptr *int2Ptr Output
int int1=26;
int int2 = 45;
int *int1Ptr = &int1;
int *int2Ptr = &int2;
*int1Ptr = 89;
*int2Ptr = 62;
int1Ptr = int2Ptr;
*int1Ptr = 80; int1 = 57;
cout << int1 <<< " "<< int2 << endl;
cout << *int1Ptr<< " "<< *int2Ptr<< endl;