Question: Without the cast operation, malloc cannot be correctly identified.

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```cpp
void Cast1() {
void *mem = malloc(100);
char *arr = (char *)mem;
arr[0] = '%'; arr[1] = '#'; arr[2] = '\0';
printf("%s\n" , arr);
free(mem);
}

/*
func.func @Cast1() attributes {llvm.linkage = #llvm.linkage} {
%c0_i8 = arith.constant 0 : i8
%c35_i8 = arith.constant 35 : i8
%c37_i8 = arith.constant 37 : i8
%c100_i64 = arith.constant 100 : i64
%0 = call @malloc(%c100_i64) : (i64) -> memref
affine.store %c37_i8, %0[0] : memref
affine.store %c35_i8, %0[1] : memref
affine.store %c0_i8, %0[2] : memref
%1 = llvm.mlir.addressof @str0 : !llvm.ptr
%2 = llvm.getelementptr %1[0, 0] : (!llvm.ptr) -> !llvm.ptr, !llvm.array<4 x i8>
%3 = "polygeist.memref2pointer"(%0) : (memref) -> !llvm.ptr
%4 = llvm.call @printf(%2, %3) vararg(!llvm.func) : (!llvm.ptr, !llvm.ptr) -> i32
memref.dealloc %0 : memref
return
}
func.func private @malloc(i64) -> memref attributes {llvm.linkage = #llvm.linkage}
*/
```

I noticed that your malloc conversion is done in VisitCastExpr, so you can only recognize malloc as a custom function if you have a special case like above.

Start by reading the malloc conversion logic in VisitCastExpr and compare it with the provided Cast1 example, where malloc is used without a cast. Trace how the call is represented and determine the expected behavior for recognizing malloc as a custom function without relying on a special cast case; done means the intended behavior is implemented or clearly documented with a regression test.

Written by the indexing model from the issue text.