The stack information of the main thread of the process cannot obtain the corresponding symbol, but the stack information of the dynamic library can obtain the corresponding symbol

[Ma1ximOne]

if (pid) {
	struct blaze_symbolize_src_process src = {
		.type_size = sizeof(src),
		.pid = pid,
	};
	syms = blaze_symbolize_process_abs_addrs(symbolizer, &src, (const uintptr_t *)stack, stack_sz);
} else {
	struct blaze_symbolize_src_kernel src = {
		.type_size = sizeof(src),
	};
	syms = blaze_symbolize_kernel_abs_addrs(symbolizer, &src, (const uintptr_t *)stack, stack_sz);
}

if (syms == NULL) {
	printf("  failed to symbolize addresses: %s\n", blaze_err_str(blaze_err_last()));
	return;
}

for (i = 0; i < stack_sz; i++) {
	if (!syms || syms->cnt <= i || syms->syms[i].name == NULL) {
		printf("%016llx: <no-symbol>\n", stack[i]);
		continue;
	}

[d-e-s-o]

I am sorry, what is the question?

[Ma1ximOne]

The user stack of the main program of the process does not have an export symbol and print < no-symbol>, the kernel stack has symbol information, but the user stack and kernel stack of the dynamic library of the process both have export symbols, what may be the reason?

[Ma1ximOne]

The process is compiled without -g