Thanks. I was able to find Thanks. I was able to find the function in curves.cpp. Its more complicated that I thought. I m guessing that gamma[0] is pwr and gamma[1] is ts. I don't know what mode is but 1 seems to work for me. void LibRaw::gamma_curve(double pwr, double ts, int mode, int imax) { int i; double g[6], bnd[2] = {0, 0}, r; g[0] = pwr; g[1] = ts; g[2] = g[3] = g[4] = 0; bnd[g[1] >= 1] = 1; if (g[1] && (g[1] - 1) * (g[0] - 1) <= 0) { for (i = 0; i < 48; i++) { g[2] = (bnd[0] + bnd[1]) / 2; if (g[0]) bnd[(pow(g[2] / g[1], -g[0]) - 1) / g[0] - 1 / g[2] > -1] = g[2]; else bnd[g[2] / exp(1 - 1 / g[2]) < g[1]] = g[2]; } g[3] = g[2] / g[1]; if (g[0]) g[4] = g[2] * (1 / g[0] - 1); } if (g[0]) g[5] = 1 / (g[1] * SQR(g[3]) / 2 - g[4] * (1 - g[3]) + (1 - pow(g[3], 1 + g[0])) * (1 + g[4]) / (1 + g[0])) - 1; else g[5] = 1 / (g[1] * SQR(g[3]) / 2 + 1 - g[2] - g[3] - g[2] * g[3] * (log(g[3]) - 1)) - 1; if (!mode--) { memcpy(gamm, g, sizeof gamm); return; } for (i = 0; i < 0x10000; i++) { curve[i] = 0xffff; if ((r = (double)i / imax) < 1) curve[i] = 0x10000 * (mode ? (r < g[3] ? r * g[1] : (g[0] ? pow(r, g[0]) * (1 + g[4]) - g[4] : log(r) * g[2] + 1)) : (r < g[2] ? r / g[1] : (g[0] ? pow((r + g[4]) / (1 + g[4]), 1 / g[0]) : exp((r - 1) / g[2])))); } } reply

Thanks. I was able to find the function in curves.cpp. Its more complicated that I thought.

I m guessing that gamma[0] is pwr and gamma[1] is ts. I don't know what mode is but 1 seems to work for me.