C- printf在for循环中打印更多线条

我在C中使用加密算法有问题，我正在编码AES Criptography，我正在调试一个函数，并且此功能打印的行要多于循环计数。在整个代码下方：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    unsigned char key[]   = {
                              0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07,
                              0x08, 0x09, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f
                             };
                              unsigned char state[] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff};
    const unsigned char sbox[256] =
    {
//0     1    2      3     4    5     6     7      8    9     A      B    C     D     E     F
        0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67, 0x2b, 0xfe, 0xd7, 0xab, 0x76, //0
        0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0, //1
        0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1, 0x71, 0xd8, 0x31, 0x15, //2
        0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a, 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75, //3
        0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29, 0xe3, 0x2f, 0x84, //4
        0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf, //5
        0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c, 0x9f, 0xa8, //6
        0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2, //7
        0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, 0x73, //8
        0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb, //9
        0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79, //A
        0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea, 0x65, 0x7a, 0xae, 0x08, //B
        0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a, //C
        0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9, 0x86, 0xc1, 0x1d, 0x9e, //D
        0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf, //E
        0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0, 0x54, 0xbb, 0x16
    }; //F
    const unsigned char Rcon[10] = {
    0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80, 0x1b, 0x36};

    unsigned char buf1, buf2, buf3, buf4, round, i;
    unsigned char rcon;

      for (round = 0; round < 10; round++){
    //add key + sbox
    for (i = 0; i <16; i++){
      state[i]=sbox[state[i] ^ key[i]];
    }
    for (i = 0; i<16; i++){
      printf("n%d",state[i]);
    }
  }
}

打印功能(如上所示(：

   for (round = 0; round < 10; round++){
        //add key + sbox
        for (i = 0; i <16; i++){
          state[i]=sbox[state[i] ^ key[i]];
        }
        for (i = 0; i<16; i++){
          printf("n%d",state[i]);
        }
      }

输出：

有人知道发生了什么事？

用循环计数器更改打印以帮助调试： -

for (i = 0; i<16; i++){
  printf("n[%d %d] %d",round, i, state[i]);
}

输出： -

[0 0] 99
[0 1] 202
....
....
[9 14] 165
[9 15] 255

10 x 16 = 160行

您有两个嵌套循环，一个带有10个，另一个迭代，因此您获得了160行印刷

(i = 0; i＆lt; 16; i (的内部循环的内部循环...打印16个数字，每行1个。外部循环的内部循环被10次调用。16行，10次为160个数字。

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