What is the value of t1 after execution of these statements if the value of t2 is as follows:```cchar t1[6];char t2[] = "Happy Studying";strncpy(t1, t2, 5);t1[5] = '\0';```Value of t1 is: [text box]**Explanation:**1. `char t1[6];` - This declares an array `t1` with space for 6 characters.2. `char t2[] = "Happy Studying";` - This initializes the array `t2` with the string "Happy Studying".3. `strncpy(t1, t2, 5);` - This copies the first 5 characters from `t2` to `t1`. After this line, `t1` becomes "Happy".4. `t1[5] = '\0';` - This adds a null terminator at the sixth position of `t1` to properly terminate the string.Therefore, the value of `t1` after these operations is: "Happy".

Lets see the solution.The value of t1 will be : Happy The value of t2 will be : Happy Studying

Answered: What is the value of t1 after execution…

What is the value of t1 after execution of these statements if the value of t2 is as follows: char t1 [ 6]; char t2 [ ]="Happy Studying" ; strncpy(t1, t2, 5); t1[5] = '\O'; Value of t1 is:

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Concept of memory addresses in pointers

The concept of pointers in high-level languages was introduced in C programming. Pointers are simply the variables that are used to store memory addresses of another variable. These special variables are generally used in the languages such as C Programming and C++ Programming, where it allows the programmer to work more deeply with memory and its addresses.

Question

What is the value of t1 after execution of these statements if the value of t2 is as follows:

```c
char t1[6];
char t2[] = "Happy Studying";
strncpy(t1, t2, 5);
t1[5] = '\0';
```

Value of t1 is: [text box]

**Explanation:**

1. `char t1[6];` - This declares an array `t1` with space for 6 characters.
2. `char t2[] = "Happy Studying";` - This initializes the array `t2` with the string "Happy Studying".
3. `strncpy(t1, t2, 5);` - This copies the first 5 characters from `t2` to `t1`. After this line, `t1` becomes "Happy".
4. `t1[5] = '\0';` - This adds a null terminator at the sixth position of `t1` to properly terminate the string.

Therefore, the value of `t1` after these operations is: "Happy".

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