C语言 如何解决分配大小为 6 的块后为 0 字节
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How to solve is 0 bytes after a block of size 6 alloc'd
提问by sinory
valgrind --tool=memcheck --leak-check=full --show-reachable=yes -v ./out
valgrind --tool=memcheck --leak-check=full --show-reachable=yes -v ./out
==37408== ERROR SUMMARY: 23 errors from 23 contexts (suppressed: 0 from 0)
==37408==
==37408== 1 errors in context 1 of 23:
==37408== Invalid read of size 1
==37408== at 0x707E: memmove$VARIANT$sse42 (in /usr/local/Cellar/valgrind/3.8.1/lib/valgrind/vgpreload_memcheck-amd64-darwin.so)
==37408== by 0x10F08F: strdup (in /usr/lib/system/libsystem_c.dylib)
==37408== by 0x10000576A: new_node (parser.c:349)
==37408== by 0x100005E5D: string_literal (parser.c:166)
==37408== by 0x100001A5D: yyparse (vtl4.y:142)
==37408== by 0x1000029F1: main (vtl4.y:218)
==37408== Address 0x100055826 is 0 bytes after a block of size 6 alloc'd
==37408== at 0x58D3: calloc (in /usr/local/Cellar/valgrind/3.8.1/lib/valgrind/vgpreload_memcheck-amd64-darwin.so)
==37408== by 0x100006734: strclip (util.c:35)
==37408== by 0x100005E4B: string_literal (parser.c:164)
==37408== by 0x100001A5D: yyparse (vtl4.y:142)
==37408== by 0x1000029F1: main (vtl4.y:218)
==37408==
==37408==
==37408== 1 errors in context 2 of 23:
==37408== Invalid read of size 1
==37408== at 0x67A4: strlen (in /usr/local/Cellar/valgrind/3.8.1/lib/valgrind/vgpreload_memcheck-amd64-darwin.so)
==37408== by 0x10F066: strdup (in /usr/lib/system/libsystem_c.dylib)
==37408== by 0x10000576A: new_node (parser.c:349)
==37408== by 0x100005E5D: string_literal (parser.c:166)
==37408== by 0x100001A5D: yyparse (vtl4.y:142)
==37408== by 0x1000029F1: main (vtl4.y:218)
==37408== Address 0x100055826 is 0 bytes after a block of size 6 alloc'd
==37408== at 0x58D3: calloc (in /usr/local/Cellar/valgrind/3.8.1/lib/valgrind/vgpreload_memcheck-amd64-darwin.so)
==37408== by 0x100006734: strclip (util.c:35)
==37408== by 0x100005E4B: string_literal (parser.c:164)
==37408== by 0x100001A5D: yyparse (vtl4.y:142)
==37408== by 0x1000029F1: main (vtl4.y:218)
==37408==
==37408== ERROR SUMMARY: 23 errors from 23 contexts (suppressed: 0 from 0)
new_node (parser.c:349)：
new_node (parser.c:349)：
struct simpleNode *new_node(JJT jjt,char *node_image){
struct simpleNode *a = malloc(sizeof(struct simpleNode));
if (a==NULL) {
yyerror("FUNC[%s]error:Init a new simple error,out of space!",__func__);
exit(0);
}
a->info.astn = jjt;
a->info.node_name = jjtNodeName[jjt];
**349**>>a->info.image = strdup(node_image);
a->info.already_rendered = cJSON_False;
a->parent = NULL;
a->firstChild = NULL;
a->nextSibling = NULL;
return a;
}
string_literal (parser.c:166):
string_literal (parser.c:166):
struct simpleNode* string_literal( char *str){
//printf("%s node!\n",__func__);
char *strc = strclip(str, strlen(str) - 2, 1);
//printf("**********strclip:%s\n",strc);
**166**>> struct simpleNode* a = new_node(JJTSTRINGLITERAL,strc);
//free(strc);
free(str);
str = NULL;
return a;
}
strclip (util.c:35)
strclip (util.c:35)
char * strclip(const char *src, unsigned long n, unsigned long m) {
unsigned long len = strlen(src);
if (n > len)
n = len - m;
if (m > len)
return NULL ;
const char * right_start = src + m;
**35**<< char *q = (char *) calloc(n, sizeof(char));
strncpy(q, right_start, n);
return q;
}
I study for a long time didn't find the root cause of the problem, please help me！
研究了半天没找到问题的根源，求大神帮帮忙！
回答by Joachim Isaksson
strncpy()in strclip()is the cause of your problem;
strncpy()instrclip()是您问题的原因；
No null-character is implicitly appended at the end of destination if source is longer than num. Thus, in this case, destination shall not be considered a null terminated C string (reading it as such would overflow).
如果 source 长于 num ，则不会在目标末尾隐式附加空字符。因此，在这种情况下，destination 不应被视为以空字符结尾的 C 字符串（这样读取它会溢出）。
Since you're using calloc()anyway to get memory filled with zeroes, all you need to do is allocate an extra char that won't be overwritten;
由于您calloc()无论如何都要使用零来填充内存，因此您需要做的就是分配一个不会被覆盖的额外字符；
char *q = (char *) calloc(n + 1, sizeof(char));
strncpy(q, right_start, n);
This will leave the last char of the buffer as zero, so you have a guaranteed termination of the string.
这将使缓冲区的最后一个字符为零，因此您可以保证字符串的终止。
