No, it does not. In C, variables have a fixed set of memory addresses to work with. If you are working on a system with 4-byte ints, and you set an int variable to 2,147,483,647 and then add 1, the variable will contain -2147483648. (On most systems. The behavior is actually undefined.) No other memory locations will be modified.

In essence, the compiler will not let you assign a value that is too big for the type. This will generate a compiler error. If you force it to with a case, the value will be truncated.

Looked at in a bitwise way, if the type can only store 8 bits, and you try to force the value 1010101010101 into it with a case, you will end up with the bottom 8 bits, or 01010101.

In your example, regardless of what you do to myArray[2], myArray[3] will contain '4'. There is no "spill over". You are trying to put something that is more than 4-bytes it will just lop off everything on the high end, leaving the bottom 4 bytes. On most systems, this will result in -2147483648.

From a practical standpoint, you want to just make sure this never, ever happens. These sorts of overflows often result in hard-to-solve defects. In other words, if you think there is any chance at all your values will be in the billions, don't use int.

No, it does not. In C, variables have a fixed set of memory addresses to work with. If you are working on a system with 4-byte ints, and you set an int variable to 2,147,483,647 and then add 1, the variable will usually contain -2147483648. (On most systems. The behavior is actually undefined.) No other memory locations will be modified.

In essence, the compiler will not let you assign a value that is too big for the type. This will generate a compiler error. If you force it to with a case, the value will be truncated.

Looked at in a bitwise way, if the type can only store 8 bits, and you try to force the value 1010101010101 into it with a case, you will end up with the bottom 8 bits, or 01010101.

In your example, regardless of what you do to myArray[2], myArray[3] will contain '4'. There is no "spill over". You are trying to put something that is more than 4-bytes it will just lop off everything on the high end, leaving the bottom 4 bytes. On most systems, this will result in -2147483648.

From a practical standpoint, you want to just make sure this never, ever happens. These sorts of overflows often result in hard-to-solve defects. In other words, if you think there is any chance at all your values will be in the billions, don't use int.

No, it does not. In C, variables have a fixed set of memory addresses to work with. If you are working on a system with 4-byte ints, and you set an int variable to 2,147,483,647 and then add 1, the variable will contain -2147483648. No other memory locations will be modified.

In essence, the compiler will not let you assign a value that is too big for the type. This will generate a compiler error. If you force it to with a case, the value will be truncated.

Looked at in a bitwise way, if the type can only store 8 bits, and you try to force the value 1010101010101 into it with a case, you will end up with the bottom 8 bits, or 01010101.

In your example, regardless of what you do to myArray[2], myArray[3] will contain '4'. There is no "spill over". You are trying to put something that is more than 4-bytes it will just lop off everything on the high end, leaving the bottom 4 bytes. On most systems, this will result in -2147483648.

From a practical standpoint, you want to just make sure this never, ever happens. These sorts of overflows often result in hard-to-solve defects. In other words, if you think there is any chance at all your values will be in the billions, don't use int.

No, it does not. In C, variables have a fixed set of memory addresses to work with. If you are working on a system with 4-byte ints, and you set an int variable to 2,147,483,647 and then add 1, the variable will contain -2147483648. (On most systems. The behavior is actually undefined.) No other memory locations will be modified.

In essence, the compiler will not let you assign a value that is too big for the type. This will generate a compiler error. If you force it to with a case, the value will be truncated.

Looked at in a bitwise way, if the type can only store 8 bits, and you try to force the value 1010101010101 into it with a case, you will end up with the bottom 8 bits, or 01010101.

In your example, regardless of what you do to myArray[2], myArray[3] will contain '4'. There is no "spill over". You are trying to put something that is more than 4-bytes it will just lop off everything on the high end, leaving the bottom 4 bytes. On most systems, this will result in -2147483648.

From a practical standpoint, you want to just make sure this never, ever happens. These sorts of overflows often result in hard-to-solve defects. In other words, if you think there is any chance at all your values will be in the billions, don't use int.