14642dc740
This change fixes two issues in ValueObject::GetExpressionPath method:
1. Accessing members of struct references used to produce expression
paths such as "str.&str.member" (instead of the expected
"str.member"). This is fixed by assigning the flag tha the child
value is a dereference when calling Dereference() on references
and adjusting logic in expression path creation.
2. If the parent of member access is dereference, the produced
expression path was "*(ptr).member". This is incorrect, since it
dereferences the member instead of the pointer. This is fixed by
wrapping dereference expression into parenthesis, resulting with
"(*(ptr)).member".
Reviewed By: werat, clayborg
Differential Revision: https://reviews.llvm.org/D132734
34 lines
566 B
C++
34 lines
566 B
C++
struct StructA {
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int x;
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int y;
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};
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struct StructB {
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int x;
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StructA &a_ref;
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StructA *&a_ptr_ref;
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};
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struct StructC : public StructB {
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int y;
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StructC(int x, StructA &a_ref, StructA *&a_ref_ptr, int y)
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: StructB{x, a_ref, a_ref_ptr}, y(y) {}
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};
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int main() {
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StructA a{1, 2};
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StructA *a_ptr = &a;
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StructB b{3, a, a_ptr};
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StructB *b_ptr = &b;
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StructB &b_ref = b;
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StructB *&b_ptr_ref = b_ptr;
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StructC c(4, a, a_ptr, 5);
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StructC *c_ptr = &c;
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StructC &c_ref = c;
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StructC *&c_ptr_ref = c_ptr;
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return 0; // Set breakpoint here
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} |