caesar | 持续集成 - 开源协议：GNU General Public License v2

First of all: You're unlikely to be able to significantly improve the performance of memcopy. Depending on the size of your buffers, where you get them from, and where the result of encoder ends up, you might be able to not copy the buffers at all and only record a reference to them, but I doubt even that will do a lot on buffers <1k.

There is one place that can be improved, though: looking at the assembler output (You have to add some non-parameterized function for actual code to be generated:

This happens because jg and jl treat the outcome of cmp as if the two operands were signed numbers.

7Ah and 8Dh represent signed numbers +122 and -115, respectively. Obviously, the latter is the smallest.

What you need is unsigned comparison. Use instructions ja and jb instead.

in uppercase condition,

keys[letter] = letters[(index + offset) % 26 + 26]

The trick here is to understand the math operations.

First of all, % is not the modulo operation, it is the remainder operation, which means that the result may be negative.

abs will create a positive value out of a negative value. But that's not the same as counteracting the problem with the negative value, as it would map e.g. -1 to 1 instead of 25.

So what you can do is to create an actual modulo operation. In Java that is:

This is because int key = (int)sKey; does NOT convert the string to an integer... at least not in the way you think it does. It takes the string pointer sKey (a memory address) to an integer. Since every time you run the progra this can be a different address, this is why it looks random. The correct way to convert a numerical string to a value is using atoi or strtol. The first part of your program should be something like this:

when you type in textInput you are searching from previous filtered data once the filtered data gets empty then all the next search will happen in empty array and you are just stuck with empty to avoid that take a dummy state which stores original data.

i have modified your code hope this helps

Since you are stating that the user input cannot be trusted, I don't recommend using the function atoi, for two reasons:

1. The behavior is undefined if the converted number is not representable as an int (i.e. if the number is out of range).

2. The function will return 0 when the conversion failed. However, when this happens, you will not know whether the user actually entered 0 or whether you got this value due to conversion failure.

Both of these problems can be solved if you use the functon strtol instead of atoi.

In your question, you state that you want to reject the input if it contains any non-digit characters. One problem with the function strtol is that it will skip any leading whitespace characters (e.g. space and tab characters) before attempting to convert a number. If you instead want to reject the input if it contains any leading whitespace characters, then you will have to compare each character with isdigit, before using the function strtol. Doing this will also solve the problem of invalid characters after the number, so that input such as 123abc will get rejected.

Here is a program which will attempt to convert argv[1] to an integer, while performing full input validation:

You forgot to add 1 to n in the test of (ord(s[n].lower())+k) < 123 so that it would count s[n] twice or more.

Change it to

I understand that at the end of every string \0 is present to mark the end of a character string.

It doesn't seem like you understand that because you didn't put such a marker at the end of shift_str but then used it as if it was a string. In your code, shift_str is not a string because there is no \0 at the end of it and therefore it is an error to pass it to printf through %s.

Perhaps change: