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https://github.com/brain-hackers/u-boot-brain
synced 2024-07-22 19:09:43 +09:00
dac326b823
In case of enable CONFIG_OF_CONTROL and has a "model" property in the root node, the board special "checkboard" will not be called. Usually we show some useful version information in the function. This patch enable call "checkboard" in any case. It is not conflicting with showing "model" at the same time. For example on LS2085AQDS: Showing "model" only: Model: Freescale Layerscape 2085a QDS Board Showing "checkboard" only: Board: LS2085E-QDS, Board Arch: V1, Board version: B, boot from vBank: 4 Showing both: Model: Freescale Layerscape 2085a QDS Board Board: LS2085E-QDS, Board Arch: V1, Board version: B, boot from vBank: 4 Signed-off-by: Haikun Wang <haikun.wang@freescale.com> Reviewed-by: Simon Glass <sjg@chromium.org>
33 lines
516 B
C
33 lines
516 B
C
/*
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* SPDX-License-Identifier: GPL-2.0+
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*/
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#include <common.h>
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#include <libfdt.h>
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#include <linux/compiler.h>
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int __weak checkboard(void)
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{
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printf("Board: Unknown\n");
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return 0;
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}
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/*
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* If the root node of the DTB has a "model" property, show it.
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* Then call checkboard().
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*/
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int show_board_info(void)
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{
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#ifdef CONFIG_OF_CONTROL
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DECLARE_GLOBAL_DATA_PTR;
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const char *model;
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model = fdt_getprop(gd->fdt_blob, 0, "model", NULL);
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if (model)
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printf("Model: %s\n", model);
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#endif
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return checkboard();
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}
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