generic_board: Call "checkboard" even though the root node has a "model" property

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>
master
Haikun.Wang@freescale.com 9 years ago committed by Tom Rini
parent 5031ca59b5
commit dac326b823
  1. 7
      common/board_info.c

@ -14,8 +14,7 @@ int __weak checkboard(void)
/*
* If the root node of the DTB has a "model" property, show it.
* If CONFIG_OF_CONTROL is disabled or the "model" property is missing,
* fall back to checkboard().
* Then call checkboard().
*/
int show_board_info(void)
{
@ -25,10 +24,8 @@ int show_board_info(void)
model = fdt_getprop(gd->fdt_blob, 0, "model", NULL);
if (model) {
if (model)
printf("Model: %s\n", model);
return 0;
}
#endif
return checkboard();

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