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What else can I help you with?
What function allocate n bytes in heap memory?
malloc
How many bytes are required to store an address of an integer?
32 bits or 4 bytes. This depends heavily on the processor architecture your computer uses, AND the programming language you are using. Typically, an integer is 1 word in length, however that processor architecture defines "word". That could be 32-bits (4 bytes), 64-bits (8 bytes), 8-bits (1 byte), or even 9 bits (in certain old computers).
How many bytes are required to store the pointer?
It depends on the platform... In a 16 bit environment, such as DOS or Windows 3.x, a near pointer is two bytes, while a far pointer is 4 bytes. In a 32 bit environment, such as Win32, a pointer is 4 bytes. In a 64 bit environment, such as Win64, a pointer is 8 bytes. If you want to find out in your particular environment, look at sizeof(ptr), where ptr is declared as a pointer to something. char* ptr; std::cout << sizeof(ptr) << std::endl; Note that the size of the pointer is not the same as the size of the object to which it points. If you looked at sizeof(*ptr), you would get 1.
What is an adequate measure of the size of input for a program that requires two integer numbers n?
Not sure what you mean; if you want to measure the "input size" in bytes, that would probably be 8 bytes, since integers typically use 4 bytes.
How an array is created in c?
For any type T, the type T[n] is an array of n Ts. If n is known at compile time, the array is fixed-length and the compiler will allocate n * sizeof(T) bytes to the array name. If n is not known at compile time, the array is variable-length and the programmer must manually request n * sizeof(T) bytes from the system at runtime, storing the start address in a pointer. The programmer must keep track of the stored address at all times and must release the memory as soon as it is no longer required.
How many bytes are required to encode n bits of data where n equals 17?
1 bytes is 8 bits so (17/8) = 2.125 so round up to 3 full bytes
How many bytes in a var-char?
In SQL, the storage size of a VARCHAR data type is determined by the length of the string stored in it, plus an additional byte (or two bytes for very large strings) to store the length of the string. Specifically, it uses 1 byte for strings up to 255 characters and 2 bytes for strings longer than 255 characters. Therefore, the total size in bytes for a VARCHAR(n) can be up to n + 1 or n + 2, depending on the length of the data.
How many bytes are required to encode n bits of data where n equals 4?
One byte is 0.125 to 1 bit. So 4 bits, is .5 bytes.
What is billion bytes on a hard disk?
On a flash drive, one billion bytes is called a(n) _____________.
How many no of address lines required in 64kB memory?
To calculate the number of address lines required for a 64 kB memory, first convert 64 kB into bytes: 64 kB = 64 × 1024 bytes = 65,536 bytes. The number of address lines needed can be determined using the formula (2^n = \text{total number of addresses}), where (n) is the number of address lines. Since 65,536 is (2^{16}), you need 16 address lines to address a 64 kB memory.
What function allocate n bytes in heap memory?
malloc
How many bytes are required to store an address of an integer?
32 bits or 4 bytes. This depends heavily on the processor architecture your computer uses, AND the programming language you are using. Typically, an integer is 1 word in length, however that processor architecture defines "word". That could be 32-bits (4 bytes), 64-bits (8 bytes), 8-bits (1 byte), or even 9 bits (in certain old computers).
How many numbers can be represented by 2 bits?
4, which is equal to 2 to the power 2.In general, with "n" bits, you can have "2 to the power n" different states (or represent that many different numbers).
How many exchange needed for reverse an array of n elements?
the integer of 1/2 n
How many bits required to store a?
To determine how many bits are required to store a specific value, you need to know the range of values that must be represented. The formula to calculate the number of bits (n) needed is ( n = \lceil \log_2(V) \rceil ), where ( V ) is the number of unique values. For example, to store integers from 0 to 255 (256 values), you would need 8 bits, since ( \log_2(256) = 8 ).
How many address lines are needed to access 256KB of main memory?
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
How many bytes are required to store the pointer?
It depends on the platform... In a 16 bit environment, such as DOS or Windows 3.x, a near pointer is two bytes, while a far pointer is 4 bytes. In a 32 bit environment, such as Win32, a pointer is 4 bytes. In a 64 bit environment, such as Win64, a pointer is 8 bytes. If you want to find out in your particular environment, look at sizeof(ptr), where ptr is declared as a pointer to something. char* ptr; std::cout << sizeof(ptr) << std::endl; Note that the size of the pointer is not the same as the size of the object to which it points. If you looked at sizeof(*ptr), you would get 1.
