1024 bytes in 1 KB,
1024 kilobyes in 1 MB
1024 x4 = 4KB x 1024 =
4194304 bytes in 1 MB. (About 4.2 million)
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A 1Mx64-bit memory chip contains <1,048,576> , or <2^20> 64-bit Lwords. You could express this either in terms of MegaBits or MegaBytes. In Megabits, 64Mbits because 1Mx64 = 64Mbits and divide by 8 to get the answer in terms of bytes. You divide by 8 because there are 8 bits in 1 byte. So a 1Mx64-bit memory contains 8 Mbytes, or 64 Mbits. If the width had been 1Mx32-bit, you would have 32Mbits, and 4 Mbytes.
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient. In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment size is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
Int: 4 bytes Float: 4 double: 8 char: 1 boolean: 1
A short is an integer that uses only 2 bytes, instead of the 4 bytes required by an int.A short is an integer that uses only 2 bytes, instead of the 4 bytes required by an int.A short is an integer that uses only 2 bytes, instead of the 4 bytes required by an int.A short is an integer that uses only 2 bytes, instead of the 4 bytes required by an int.
The storage size of an int in C is loosely defined, and may be either 2 bytes or, more commonly, 4 bytes. Whether or not it is defined as const won't affect the size.