ARM: imx: initialize clk_init_data.flags for clk-fixup-mux
The clk_init_data.flags of clk-fixup-mux is left there without initialization. It may hold some random data and cause clock framework interpret the clock in an unexpected way. At least on imx6sl, the following division by zero error with sched_clock is seen because of it. Division by zero in kernel. CPU: 0 PID: 0 Comm: swapper/0 Not tainted 3.11.0-rc3+ #19 Backtrace: [<80011af0>] (dump_backtrace+0x0/0x10c) from [<80011c90>] (show_stack+0x18/0x1c) r6:3b9aca00 r5:00000020 r4:00000000 r3:00000000 [<80011c78>] (show_stack+0x0/0x1c) from [<8055e02c>] (dump_stack+0x78/0x94) [<8055dfb4>] (dump_stack+0x0/0x94) from [<80011924>] (__div0+0x18/0x20) r4:00000000 r3:00000000 [<8001190c>] (__div0+0x0/0x20) from [<8026c408>] (Ldiv0_64+0x8/0x18) [<8006330c>] (clocks_calc_mult_shift+0x0/0xf8) from [<8072f604>] (setup_sched_clock+0x88/0x1f0) [<8072f57c>] (setup_sched_clock+0x0/0x1f0) from [<8071ad48>] (mxc_timer_init+0xe8/0x17c) [<8071ac60>] (mxc_timer_init+0x0/0x17c) from [<807290b0>] (imx6sl_clocks_init+0x1db8/0x1dc0) r8:807a9ca4 r7:00000000 r6:80777564 r5:8100c1f4 r4:c0820000 [<807272f8>] (imx6sl_clocks_init+0x0/0x1dc0) from [<807420ac>] (of_clk_init+0x40/0x6c) [<8074206c>] (of_clk_init+0x0/0x6c) from [<807290cc>] (imx6sl_timer_init+0x14/0x18) r5:807a8e80 r4:ffffffff [<807290b8>] (imx6sl_timer_init+0x0/0x18) from [<80716e1c>] (time_init+0x24/0x34) [<80716df8>] (time_init+0x0/0x34) from [<80713738>] (start_kernel+0x1b0/0x310) [<80713588>] (start_kernel+0x0/0x310) from [<80008074>] (0x80008074) r7:80770b08 r6:80754cd4 r5:8076c8c4 r4:10c53c7d sched_clock: 32 bits at 0 Hz, resolution 0ns, wraps every 0ms Fix the bug by initializing init.flags as zero. Signed-off-by: Shawn Guo <shawn.guo@linaro.org>
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@ -90,6 +90,7 @@ struct clk *imx_clk_fixup_mux(const char *name, void __iomem *reg,
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init.ops = &clk_fixup_mux_ops;
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init.parent_names = parents;
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init.num_parents = num_parents;
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init.flags = 0;
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fixup_mux->mux.reg = reg;
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fixup_mux->mux.shift = shift;
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