运行时检查失败 #3 变量 'i1' 正在使用而未初始化

嗨，我是该网站和 c++11 的新手，我尝试对此进行调查，但我似乎收到以下错误，"运行时检查失败 #3 变量'i1'在没有初始化的情况下被使用" 我认为这正是错误所说的，我没有初始化变量，但我一遍又一遍地检查，似乎无法发现为什么它不是已初始化。希望有人能帮忙。这是我的代码：

void cipher(array<char, array_rows>& text1, array<char, array_rows>& text2, array<int, 52>& key)
{
int i1=0;                 // Index into first text array.
int i2=0;                 // Index into second text array.
int ik=0;                     // Index into key array.
int x1=0;
int x2=0;
int x3=0;   
int x4=0; 
int t1=0; 
int t2=0; // Four "16-bit" blocks, two temps.
int r=0;                      // Eight rounds of processing.
int i=0;
auto num_threads = thread::hardware_concurrency();

#pragma omp  parallel  num_threads(num_threads) default(none) shared(text1, text2, key) private(i, i1,i2,ik,x1,x2,x3,x4,t1,t2,r)  
//#pragma for <----- this commented or not gives same error 
for ( i = 0; i < text1.size(); i += 8)
{
    ik = 0;                 // Restart key index.
    r = 8;                  // Eight rounds of processing.
    // Load eight plain1 bytes as four 16-bit "unsigned" integers.
    // Masking with 0xff prevents sign extension with cast to int.
    x1 = text1[i1++] & 0xff;          // Build 16-bit x1 from 2 bytes,
    x1 |= (text1[i1++] & 0xff) << 8;  // assuming low-order byte first.
    x2 = text1[i1++] & 0xff;
    x2 |= (text1[i1++] & 0xff) << 8;
    x3 = text1[i1++] & 0xff;
    x3 |= (text1[i1++] & 0xff) << 8;
    x4 = text1[i1++] & 0xff;
    x4 |= (text1[i1++] & 0xff) << 8;
    do 
    {
        // 1) Multiply (modulo 0x10001), 1st text sub-block
        // with 1st key sub-block.
        x1 = (int) ((long long) x1 * key[ik++] % 0x10001L & 0xffff);
        // 2) Add (modulo 0x10000), 2nd text sub-block
        // with 2nd key sub-block.
        x2 = x2 + key[ik++] & 0xffff;
        // 3) Add (modulo 0x10000), 3rd text sub-block
        // with 3rd key sub-block.
        x3 = x3 + key[ik++] & 0xffff;
        // 4) Multiply (modulo 0x10001), 4th text sub-block
        // with 4th key sub-block.
        x4 = (int) ((long long) x4 * key[ik++] % 0x10001L & 0xffff);
        // 5) XOR results from steps 1 and 3.
        t2 = x1 ^ x3;
        // 6) XOR results from steps 2 and 4.
        // Included in step 8.
        // 7) Multiply (modulo 0x10001), result of step 5
        // with 5th key sub-block.
        t2 = (int) ((long long) t2 * key[ik++] % 0x10001L & 0xffff);
        // 8) Add (modulo 0x10000), results of steps 6 and 7.
        t1 = t2 + (x2 ^ x4) & 0xffff;
        // 9) Multiply (modulo 0x10001), result of step 8
        // with 6th key sub-block.
        t1 = (int) ((long long) t1 * key[ik++] % 0x10001L & 0xffff);
        // 10) Add (modulo 0x10000), results of steps 7 and 9.
        t2 = t1 + t2 & 0xffff;
        // 11) XOR results from steps 1 and 9.
        x1 ^= t1;
        // 14) XOR results from steps 4 and 10. (Out of order).
        x4 ^= t2;
        // 13) XOR results from steps 2 and 10. (Out of order).
        t2 ^= x2;
        // 12) XOR results from steps 3 and 9. (Out of order).
        x2 = x3 ^ t1;
        x3 = t2;        // Results of x2 and x3 now swapped.
    } while(--r != 0);  // Repeats seven more rounds.
    // Final output transform (4 steps).
    // 1) Multiply (modulo 0x10001), 1st text-block
    // with 1st key sub-block.
    x1 = (int) ((long long) x1 * key[ik++] % 0x10001L & 0xffff);
    // 2) Add (modulo 0x10000), 2nd text sub-block
    // with 2nd key sub-block. It says x3, but that is to undo swap
    // of subblocks 2 and 3 in 8th processing round.
    x3 = x3 + key[ik++] & 0xffff;
    // 3) Add (modulo 0x10000), 3rd text sub-block
    // with 3rd key sub-block. It says x2, but that is to undo swap
    // of subblocks 2 and 3 in 8th processing round.
    x2 = x2 + key[ik++] & 0xffff;
    // 4) Multiply (modulo 0x10001), 4th text-block
    // with 4th key sub-block.
    x4 = (int) ((long long) x4 * key[ik++] % 0x10001L & 0xffff);
    // Repackage from 16-bit sub-blocks to 8-bit byte array text2.
    text2[i2++] = (char)x1;
    text2[i2++] = (char)(x1 >> 8);
    text2[i2++] = (char)x3;                // x3 and x2 are switched
    text2[i2++] = (char)(x3 >> 8);        // only in name.
    text2[i2++] = (char)x2;
    text2[i2++] = (char)(x2 >> 8);
    text2[i2++] = (char)x4;
    text2[i2++] = (char)(x4 >> 8);
}   // End for loop.
}