与COCCINELLE的malloc之后将缺少null检查
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19-09-2019 - |
题
我想写 COCCINELLE 语义补丁,这样它会调用之后添加if (ptr == NULL) ...
检查的malloc在那里它们被丢失。
让我们说我有下列输入源代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// memory leaks ignored
static void OK_simple(void)
{
char *ptr;
ptr = malloc(100);
if (ptr == NULL) {
return;
}
strcpy(ptr, "abcd");
}
static void NOT_OK_missing_NULL_check(void)
{
char *ptr;
ptr = malloc(100);
strcpy(ptr, "abcd");
}
static void NOT_OK_NULL_check_too_late(void)
{
char *ptr;
ptr = malloc(100);
strcpy(ptr, "abcd");
if (ptr == NULL) {
return;
}
}
static void OK_code_between_allocation_and_NULL_check(void)
{
char *ptr;
ptr = malloc(100);
printf("The NULL test does not have to be put immediately after\n");
if (ptr == NULL) {
return;
}
strcpy(ptr, "abcd");
}
static void OK_two_allocations(void)
{
char *ptr1, *ptr2;
ptr1 = malloc(100);
ptr2 = malloc(100);
if (ptr1 == NULL) {
return;
}
if (ptr2 == NULL) {
return;
}
strcpy(ptr1, "abcd");
strcpy(ptr2, "abcd");
}
static void NOT_OK_two_allocations_with_one_missing_NULL_check(void)
{
char *ptr1, *ptr2;
ptr1 = malloc(100);
ptr2 = malloc(100);
if (ptr1 == NULL) {
return;
}
strcpy(ptr1, "abcd");
strcpy(ptr2, "abcd");
}
int main(int argc, char *argv[])
{
(void)argc;
(void)argv;
OK_simple();
NOT_OK_missing_NULL_check();
NOT_OK_NULL_check_too_late();
OK_code_between_allocation_and_NULL_check();
OK_two_allocations();
NOT_OK_two_allocations_with_one_missing_NULL_check();
return 0;
}
我一直在试图拿出一个语义补丁,它这样做,但我有麻烦它做我想做的。只需添加NULL测试无条件地是没有问题的,麻烦的是不这样做时,没有必要。以下是我目前有:
// this rule matches code that already have a NULL test
@already_have_proper_check@
statement S;
type T;
T* ptr;
expression E;
@@
ptr = malloc(E);
... when != ptr
if (ptr == NULL) S
//+dummy_change_just_to_verify_that_this_rule_matches();
// this rule adds NULL tests where missing
//@add_NULL_check depends on !already_have_proper_check@
@add_NULL_check@
type T;
//T* ptr != already_have_proper_check.ptr;
T* ptr;
expression E;
@@
ptr = malloc(E);
+if (ptr == NULL) {
+ insert_error_handling_here();
+}
这个麻烦的是,它并不排除确定的情况下,我不知道如何将两个规则连接。谁能帮我这个?
只要为100%清楚,我跑COCCINELLE后要输出的是以下内容:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// memory leaks ignored
static void OK_simple(void)
{
char *ptr;
ptr = malloc(100);
if (ptr == NULL) {
return;
}
strcpy(ptr, "abcd");
}
static void NOT_OK_missing_NULL_check(void)
{
char *ptr;
ptr = malloc(100);
if (ptr == NULL) {
insert_error_handling_here();
}
strcpy(ptr, "abcd");
}
static void NOT_OK_NULL_check_too_late(void)
{
char *ptr;
ptr = malloc(100);
if (ptr == NULL) {
insert_error_handling_here();
}
strcpy(ptr, "abcd");
if (ptr == NULL) {
return;
}
}
static void OK_code_between_allocation_and_NULL_check(void)
{
char *ptr;
ptr = malloc(100);
printf("The NULL test does not have to be put immediately after\n");
if (ptr == NULL) {
return;
}
strcpy(ptr, "abcd");
}
static void OK_two_allocations(void)
{
char *ptr1, *ptr2;
ptr1 = malloc(100);
ptr2 = malloc(100);
if (ptr1 == NULL) {
return;
}
if (ptr2 == NULL) {
return;
}
strcpy(ptr1, "abcd");
strcpy(ptr2, "abcd");
}
static void NOT_OK_two_allocations_with_one_missing_NULL_check(void)
{
char *ptr1, *ptr2;
ptr1 = malloc(100);
ptr2 = malloc(100);
if (ptr2 == NULL) {
insert_error_handling_here();
}
if (ptr1 == NULL) {
return;
}
strcpy(ptr1, "abcd");
strcpy(ptr2, "abcd");
}
int main(int argc, char *argv[])
{
(void)argc;
(void)argv;
OK_simple();
NOT_OK_missing_NULL_check();
NOT_OK_NULL_check_too_late();
OK_code_between_allocation_and_NULL_check();
OK_two_allocations();
NOT_OK_two_allocations_with_one_missing_NULL_check();
return 0;
}
解决方案
// find calls to malloc @call@ expression ptr; position p; @@ ptr@p = malloc(...); // find ok calls to malloc @ok@ expression ptr; position call.p; @@ ptr@p = malloc(...); ... when != ptr ( (ptr == NULL || ...) | (ptr != NULL || ...) ) // fix bad calls to malloc @depends on !ok@ expression ptr; position call.p; @@ ptr@p = malloc(...); + if (ptr == NULL) return;
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