1. 程式人生 > >xcode編譯時由於函式重複導致的問題

xcode編譯時由於函式重複導致的問題

xcode編譯時由於函式重複導致的問題

polarssllibtomcrypt某個函式衝突的問題

現象:

說明:

我的程式分為主程式和sdk程式(framework),sdk中包括polarssl庫(原始檔)、libtomcrypt庫(.a靜態庫檔案)

問題一:

當都在framework中新增lib,在主程式中不新增libtomcrypt,則在使用polarssl庫的ssl握手的時候會出現ssl握手失敗:返回-30592

問題二:

當在主程式中也新增libtomcrypt.a庫,則ssl握手會成功,但是在執行中的某個步驟會報錯:hash_descriptor[hash_idx].process(

&md, pInData,InDataLen)err

解決方法:

由於兩個庫中有相同的函式(函式名、函式引數都相同),所以導致兩個庫的方法串了;

我的polarssl是原始檔,將函式名改掉就好(如:md5_process  —改為—> md5_process1),h檔案和c檔案都要改掉,再執行就不會出現上述問題了

對比一下重複的內容吧:

polarssl中的md5_process方法:

void md5_process1( md5_context *ctx, const unsigned char data[64] )
{
    uint32_t X[16], A, B, C, D;

    GET_UINT32_LE( X[ 0], data,  0 );
    GET_UINT32_LE( X[ 1], data,  4 );
    GET_UINT32_LE( X[ 2], data,  8 );
    GET_UINT32_LE( X[ 3], data, 12 );
    GET_UINT32_LE( X[ 4], data, 16 );
    GET_UINT32_LE( X[ 5], data, 20 );
    GET_UINT32_LE( X[ 6], data, 24 );
    GET_UINT32_LE( X[ 7], data, 28 );
    GET_UINT32_LE( X[ 8], data, 32 );
    GET_UINT32_LE( X[ 9], data, 36 );
    GET_UINT32_LE( X[10], data, 40 );
    GET_UINT32_LE( X[11], data, 44 );
    GET_UINT32_LE( X[12], data, 48 );
    GET_UINT32_LE( X[13], data, 52 );
    GET_UINT32_LE( X[14], data, 56 );
    GET_UINT32_LE( X[15], data, 60 );

#define S(x,n) ((x << n) | ((x & 0xFFFFFFFF) >> (32 - n)))

#define P(a,b,c,d,k,s,t)                                \
{                                                       \
    a += F(b,c,d) + X[k] + t; a = S(a,s) + b;           \
}

    A = ctx->state[0];
    B = ctx->state[1];
    C = ctx->state[2];
    D = ctx->state[3];

#define F(x,y,z) (z ^ (x & (y ^ z)))

    P( A, B, C, D,  0,  7, 0xD76AA478 );
    P( D, A, B, C,  1, 12, 0xE8C7B756 );
    P( C, D, A, B,  2, 17, 0x242070DB );
    P( B, C, D, A,  3, 22, 0xC1BDCEEE );
    P( A, B, C, D,  4,  7, 0xF57C0FAF );
    P( D, A, B, C,  5, 12, 0x4787C62A );
    P( C, D, A, B,  6, 17, 0xA8304613 );
    P( B, C, D, A,  7, 22, 0xFD469501 );
    P( A, B, C, D,  8,  7, 0x698098D8 );
    P( D, A, B, C,  9, 12, 0x8B44F7AF );
    P( C, D, A, B, 10, 17, 0xFFFF5BB1 );
    P( B, C, D, A, 11, 22, 0x895CD7BE );
    P( A, B, C, D, 12,  7, 0x6B901122 );
    P( D, A, B, C, 13, 12, 0xFD987193 );
    P( C, D, A, B, 14, 17, 0xA679438E );
    P( B, C, D, A, 15, 22, 0x49B40821 );

#undef F

#define F(x,y,z) (y ^ (z & (x ^ y)))

    P( A, B, C, D,  1,  5, 0xF61E2562 );
    P( D, A, B, C,  6,  9, 0xC040B340 );
    P( C, D, A, B, 11, 14, 0x265E5A51 );
    P( B, C, D, A,  0, 20, 0xE9B6C7AA );
    P( A, B, C, D,  5,  5, 0xD62F105D );
    P( D, A, B, C, 10,  9, 0x02441453 );
    P( C, D, A, B, 15, 14, 0xD8A1E681 );
    P( B, C, D, A,  4, 20, 0xE7D3FBC8 );
    P( A, B, C, D,  9,  5, 0x21E1CDE6 );
    P( D, A, B, C, 14,  9, 0xC33707D6 );
    P( C, D, A, B,  3, 14, 0xF4D50D87 );
    P( B, C, D, A,  8, 20, 0x455A14ED );
    P( A, B, C, D, 13,  5, 0xA9E3E905 );
    P( D, A, B, C,  2,  9, 0xFCEFA3F8 );
    P( C, D, A, B,  7, 14, 0x676F02D9 );
    P( B, C, D, A, 12, 20, 0x8D2A4C8A );

#undef F
    
#define F(x,y,z) (x ^ y ^ z)

    P( A, B, C, D,  5,  4, 0xFFFA3942 );
    P( D, A, B, C,  8, 11, 0x8771F681 );
    P( C, D, A, B, 11, 16, 0x6D9D6122 );
    P( B, C, D, A, 14, 23, 0xFDE5380C );
    P( A, B, C, D,  1,  4, 0xA4BEEA44 );
    P( D, A, B, C,  4, 11, 0x4BDECFA9 );
    P( C, D, A, B,  7, 16, 0xF6BB4B60 );
    P( B, C, D, A, 10, 23, 0xBEBFBC70 );
    P( A, B, C, D, 13,  4, 0x289B7EC6 );
    P( D, A, B, C,  0, 11, 0xEAA127FA );
    P( C, D, A, B,  3, 16, 0xD4EF3085 );
    P( B, C, D, A,  6, 23, 0x04881D05 );
    P( A, B, C, D,  9,  4, 0xD9D4D039 );
    P( D, A, B, C, 12, 11, 0xE6DB99E5 );
    P( C, D, A, B, 15, 16, 0x1FA27CF8 );
    P( B, C, D, A,  2, 23, 0xC4AC5665 );

#undef F

#define F(x,y,z) (y ^ (x | ~z))

    P( A, B, C, D,  0,  6, 0xF4292244 );
    P( D, A, B, C,  7, 10, 0x432AFF97 );
    P( C, D, A, B, 14, 15, 0xAB9423A7 );
    P( B, C, D, A,  5, 21, 0xFC93A039 );
    P( A, B, C, D, 12,  6, 0x655B59C3 );
    P( D, A, B, C,  3, 10, 0x8F0CCC92 );
    P( C, D, A, B, 10, 15, 0xFFEFF47D );
    P( B, C, D, A,  1, 21, 0x85845DD1 );
    P( A, B, C, D,  8,  6, 0x6FA87E4F );
    P( D, A, B, C, 15, 10, 0xFE2CE6E0 );
    P( C, D, A, B,  6, 15, 0xA3014314 );
    P( B, C, D, A, 13, 21, 0x4E0811A1 );
    P( A, B, C, D,  4,  6, 0xF7537E82 );
    P( D, A, B, C, 11, 10, 0xBD3AF235 );
    P( C, D, A, B,  2, 15, 0x2AD7D2BB );
    P( B, C, D, A,  9, 21, 0xEB86D391 );

#undef F

    ctx->state[0] += A;
    ctx->state[1] += B;
    ctx->state[2] += C;
    ctx->state[3] += D;
}


libtomcrypt庫的md5_process方法不是直接顯示的,分為兩個部分,1呼叫,2巨集定義

呼叫程式碼如下:

/**
   Process a block of memory though the hash
   @param md     The hash state
   @param in     The data to hash
   @param inlen  The length of the data (octets)
   @return CRYPT_OK if successful
*/
HASH_PROCESS(md5_process, md5_compress, md5, 64)


巨集定義如下:

/* a simple macro for making hash "process" functions */
#define HASH_PROCESS(func_name, compress_name, state_var, block_size)                       \
int func_name (hash_state * md, const unsigned char *in, unsigned long inlen)               \
{                                                                                           \
    unsigned long n;                                                                        \
    int           err;                                                                      \
    LTC_ARGCHK(md != NULL);                                                                 \
    LTC_ARGCHK(in != NULL);                                                                 \
    if (md-> state_var .curlen > sizeof(md-> state_var .buf)) {                             \
       return CRYPT_INVALID_ARG;                                                            \
    }                                                                                       \
    while (inlen > 0) {                                                                     \
        if (md-> state_var .curlen == 0 && inlen >= block_size) {                           \
           if ((err = compress_name (md, (unsigned char *)in)) != CRYPT_OK) {               \
              return err;                                                                   \
           }                                                                                \
           md-> state_var .length += block_size * 8;                                        \
           in             += block_size;                                                    \
           inlen          -= block_size;                                                    \
        } else {                                                                            \
           n = MIN(inlen, (block_size - md-> state_var .curlen));                           \
           memcpy(md-> state_var .buf + md-> state_var.curlen, in, (size_t)n);              \
           md-> state_var .curlen += n;                                                     \
           in             += n;                                                             \
           inlen          -= n;                                                             \
           if (md-> state_var .curlen == block_size) {                                      \
              if ((err = compress_name (md, md-> state_var .buf)) != CRYPT_OK) {            \
                 return err;                                                                \
              }                                                                             \
              md-> state_var .length += 8*block_size;                                       \
              md-> state_var .curlen = 0;                                                   \
           }                                                                                \
       }                                                                                    \
    }                                                                                       \
    return CRYPT_OK;                                                                        \
}



解決過程:

過程很累很複雜,也走過一些彎路。。

1.猜測兩個庫中有相同的函式(函式名、函式引數都相同)

1)嘗試xcode能否按路徑匯入庫,就不會將函式串了:①在標頭檔案上面下功夫,嘗試includeimpor<>和””;②把libtomcrypt.a庫加到主程式,導致問題二;——————失敗

2)嘗試開啟polarssldebug,一步一步的看握手過程到底哪裡出了問題,由於不是當前步驟報錯,是伺服器返回一個fatal message,可以判斷是之前的某個步驟內容設定問題,但是不能判斷到底是哪個步驟,失敗了

2.猜測libtomcrypt.a庫版本不對,但是很明顯不是這個問題

過程太複雜,不記得了,想起來再說吧。。