帮我分析一下以下代码:char Source[264]; // [esp+D0h] [ebp-144h] BYREF unsigned int v3; // [esp+1D8h] [ebp-3Ch] unsigned int v4; // [esp+1E4h] [ebp-30h] size_t v5; // [esp+1F0h] [ebp-24h] unsigned int i; // [esp+1FCh] [ebp-18h] int v7; // [esp+208h] [ebp-Ch] __CheckForDebuggerJustMyCode(&unk_BA3015); j__memset(Source, 0, 0x100u); v5 = j__strlen(Destination); v4 = v5 / 3; v3 = v5 % 3; v7 = 0; for ( i = 0; i < 3 * v4; i += 3 ) { Source[v7] = dword_B9F000 + aAbcdefghijklmn[(Destination[i] & 0xFC) >> 2]; Source[v7 + 1] = dword_B9F000 + aAbcdefghijklmn[(Destination[i + 1] >> 4) | (16 * (Destination[i] & 3))]; Source[v7 + 2] = dword_B9F000 + aAbcdefghijklmn[((Destination[i + 2] & 0xC0) >> 6) | (4 * (Destination[i + 1] & 0xF))]; Source[v7 + 3] = dword_B9F000 + aAbcdefghijklmn[Destination[i + 2] & 0x3F]; v7 += 4; } if ( v3 == 2 ) { Source[4 * v4] = dword_B9F000 + aAbcdefghijklmn[(Destination[3 * v4] & 0xFC) >> 2]; Source[4 * v4 + 1] = dword_B9F000 + aAbcdefghijklmn[(Destination[3 * v4 + 1] >> 4) | (16 * (Destination[3 * v4] & 3))]; Source[4 * v4 + 2] = dword_B9F000 + aAbcdefghijklmn[4 * (Destination[3 * v4 + 1] & 0xF)]; Source[4 * v4 + 3] = 64; } if ( v3 == 1 ) { Source[4 * v4] = dword_B9F000 + aAbcdefghijklmn[(Destination[3 * v4] & 0xFC) >> 2]; Source[4 * v4 + 1] = dword_B9F000 + aAbcdefghijklmn[16 * (Destination[3 * v4] & 3)]; Source[4 * v4 + 2] = 64; Source[4 * v4 + 3] = 35; } return j__strcpy_s(Destination, 0x100u, Source);
时间: 2023-03-24 11:05:03 浏览: 113
我写的代码,谁能帮我解决问题
3星 · 编辑精心推荐
这段代码定义了一个名为 Source 的字符数组,大小为 264。同时定义了四个变量 v3、v4、v5 和 i,分别为无符号整型和 size_t 类型。最后定义了一个整型变量 v7。这段代码中还有一个函数调用 __CheckForDebuggerJustMyCode(),该函数用于检测是否有调试器正在运行。
阅读全文