数组越界添加字节数组到另一个

Question

进行一点加密工作，并遇到了超出范围的异常数组。 我在纸上找了几次，对我来说一切都还好，所以我不能真正确定错误的起因。 如果有人可以帮助，那就太好了！

   static byte[] encrypt(byte[] ptBytes, javax.crypto.SecretKey key, byte[] IV){

    byte [] ct; 
    byte [] pt;
    byte [] ptBlock, ctBlock;
    int bytes;              //the number of bytes left over in the last block // this comes into play w/ the last 2 blocks witht the swap and stuff

    //get the extra bytes in case last block of plain text isn't whole
    bytes = ptBytes.length%16;

    //pad the plain text array to proper length
    pt = Arrays.copyOf(ptBytes, (((ptBytes.length )/16) + 1) * 16 );
    System.out.println(Arrays.toString(pt));

    //ctBlock = one block of cipher text
    ctBlock = new byte [16];

    //make ct the length of the padded pt 
    ct = new byte [pt.length];

    //do the encryption
    //i is for the current block of plain / cipher text we are on
    for( int i = 1; i <= ((ptBytes.length )/16)+1; i++){
        if( i == 1 ){

            //make ptBlock the first block of the entire plain text
            ptBlock = Arrays.copyOfRange(pt, 0, (i*16));

            //since i = 1 do the XOR to get new plain text with IV
            for (int j = 0; j < ptBlock.length - 1; j++){
                ptBlock[j] = (byte)(ptBlock[j] ^ IV[j]);
            }

            //now time to do the encryption between the current block of plain text and the key
            try {
                ctBlock = AES.encrypt(ptBlock, key);
            } catch (Exception e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            //now put the cipher text block we just got into the final cipher text array
            for( int k = 0; k < ctBlock.length; k++){
                ct[k] = ctBlock[k];
            }


        }
        else{
            //make ptBlock the current number block of entire plain text
            ptBlock = Arrays.copyOfRange(pt, (i-1)*16, (i*16));

            //now XOR the plain text block with the prior cipher text block
            for(int j = 0; j < ptBlock.length - 1; j++){
                ptBlock[i] = (byte) (ptBlock[j] ^ ctBlock[j]);
            }

            //now time to do the encryption between the current block of plain text and the key
            try {
                ctBlock = AES.encrypt(ptBlock, key);
            } catch (Exception e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            //now put the cipher text block we just got into the final cipher text array
            for( int k = (i-1)*16; k < (i*16)-1; k++){
                ct[k] = ctBlock[k-16];
            }
        }
    }

    return ct;
}

它说错误在这条线上

ct[k] = ctBlock[k-16];

这没有多大意义。 数组ct的长度为48，而ctBlock的长度为len 16，并且在for循环中出现此错误的情况下，i等于2或3，因此我将一个16字节大小的数组添加到数组的第二个三分之一数组ct或第3个三分之一。 就像我说的，我在纸上找到了它，这似乎很合法！

提前致谢！

Answer 1

考虑当i = 3时的情况。

for( int k = (i-1)*16; k < (i*16)-1; k++){
    ct[k] = ctBlock[k-16];
}

这里 -

• k从32开始
• 条件变为32 <47
• ctBlock数组索引变为ctBlock = 16，而且很糟糕！ 数组索引超出范围！

快速解决 -

for( int k = (i - 1) * 16; k < (i * 16) - 1; k++){
    ct[k] = ctBlock[k - (16 * (i - 1))];
}

Answer 2

如果i == 3 ，如您所说，则k == 32开始，并增加到47 。 您说ctBlock是一个大小为16的数组，在这种情况下，尝试访问索引16到31处的元素将引发错误。