imx-drm: ipu-dp: Introduce IPUV3_NUM_FLOWS
IPUv3 has a total of 3 flows (one synchronous flow and 2 asynchronous flows). Let's add a definition for such number in order to let the code easier to understand. Signed-off-by: Fabio Estevam <fabio.estevam@freescale.com> Acked-by: Sascha Hauer <s.hauer@pengutronix.de> Acked-by: Philipp Zabel <p.zabel@pengutronix.de> Signed-off-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>
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1 changed files with 5 additions and 3 deletions
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@ -46,6 +46,8 @@
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#define DP_COM_CONF_CSC_DEF_BG (2 << 8)
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#define DP_COM_CONF_CSC_DEF_BOTH (1 << 8)
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#define IPUV3_NUM_FLOWS 3
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struct ipu_dp_priv;
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struct ipu_dp {
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@ -67,7 +69,7 @@ struct ipu_dp_priv {
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struct ipu_soc *ipu;
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struct device *dev;
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void __iomem *base;
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struct ipu_flow flow[3];
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struct ipu_flow flow[IPUV3_NUM_FLOWS];
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struct mutex mutex;
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int use_count;
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};
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@ -280,7 +282,7 @@ struct ipu_dp *ipu_dp_get(struct ipu_soc *ipu, unsigned int flow)
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struct ipu_dp_priv *priv = ipu->dp_priv;
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struct ipu_dp *dp;
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if (flow > 5)
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if ((flow >> 1) >= IPUV3_NUM_FLOWS)
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return ERR_PTR(-EINVAL);
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if (flow & 1)
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@ -322,7 +324,7 @@ int ipu_dp_init(struct ipu_soc *ipu, struct device *dev, unsigned long base)
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mutex_init(&priv->mutex);
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for (i = 0; i < 3; i++) {
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for (i = 0; i < IPUV3_NUM_FLOWS; i++) {
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priv->flow[i].foreground.foreground = 1;
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priv->flow[i].base = priv->base + ipu_dp_flow_base[i];
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priv->flow[i].priv = priv;
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