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e2c3142274
If F is a Thumb function symbol, and G = F + const, and G is a function symbol, then G is Thumb. Because what else could it be? Differential Revision: https://reviews.llvm.org/D28878 git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@292514 91177308-0d34-0410-b5e6-96231b3b80d8
41 lines
880 B
ArmAsm
41 lines
880 B
ArmAsm
@@ test st_value bit 0 of thumb function
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@ RUN: llvm-mc %s -triple=thumbv7-linux-gnueabi -filetype=obj -o - | \
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@ RUN: llvm-readobj -t | FileCheck %s
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.syntax unified
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.text
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.globl foo
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.align 2
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.code 16
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.thumb_func
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.type foo,%function
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foo:
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bx lr
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.global bar
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bar = foo
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.global baz
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baz = foo + 2
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@@ make sure foo and bar are thumb function: bit 0 = 1 (st_value)
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@CHECK: Symbol {
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@CHECK: Name: bar
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@CHECK-NEXT: Value: 0x1
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@CHECK-NEXT: Size: 0
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@CHECK-NEXT: Binding: Global
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@CHECK-NEXT: Type: Function
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@CHECK: Symbol {
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@CHECK: Name: baz
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@CHECK-NEXT: Value: 0x3
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@CHECK-NEXT: Size: 0
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@CHECK-NEXT: Binding: Global
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@CHECK-NEXT: Type: Function
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@CHECK: Symbol {
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@CHECK: Name: foo
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@CHECK-NEXT: Value: 0x1
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@CHECK-NEXT: Size: 0
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@CHECK-NEXT: Binding: Global
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@CHECK-NEXT: Type: Function
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