2016-05-23 151 views
0

我正在尝试阅读json字符串并打印,它正在生成段错误(核心转储)。我认为错误是输入字符串,但不是很确定。分段错误读取json字符串

这里是代码

CODE:

#include <json/json.h> 
#include <stdio.h> 

void json_parse(json_object * jobj); 

int main(){ 

char * string = "{" 
       "\"coooooool\": { " 
            "\"name\" : \"coooooooooool\"," 
            "\"name\" : 1" 
            "\"}" 
       "\"}"; 

printf ("JSON string: %sn", string); 
json_object * jobj = json_tokener_parse(string); 
json_parse(jobj); 


return 0; 
} 

void json_parse(json_object * jobj) { 
enum json_type type; 
json_object_object_foreach(jobj, key, val) { 
    type = json_object_get_type(val); 
    switch (type) 
    { 
     case json_type_int: printf("type: json_type_int, "); 
     printf("value: %dn", json_object_get_int(val)); 
     break; 
    } 
    } 
} 

我跑了Valgrind的输出二进制文件,检查错误正确

我得到这个误差的valgrind运行时

==14573== Memcheck, a memory error detector 
==14573== Copyright (C) 2002-2011, and GNU GPL'd, by Julian Seward et al. 
==14573== Using Valgrind-3.7.0 and LibVEX; rerun with -h for copyright info 
==14573== Command: ./a.out 
==14573== 
==14573== Invalid read of size 4 
==14573== at 0x40491F8: json_object_get_object (in /usr/lib/i386-linux-gnu/libjson.so.0.0.1) 
==14573== by 0x80485E5: main (in /var/www/json/a.out) 
==14573== Address 0xfffffff5 is not stack'd, malloc'd or (recently) free'd 
==14573== 
==14573== 
==14573== Process terminating with default action of signal 11 (SIGSEGV) 
==14573== Access not within mapped region at address 0xFFFFFFF5 
==14573== at 0x40491F8: json_object_get_object (in /usr/lib/i386-linux-gnu/libjson.so.0.0.1) 
==14573== by 0x80485E5: main (in /var/www/json/a.out) 
==14573== If you believe this happened as a result of a stack 
==14573== overflow in your program's main thread (unlikely but 
==14573== possible), you can try to increase the size of the 
==14573== main thread stack using the --main-stacksize= flag. 
==14573== The main thread stack size used in this run was 8388608. 
JSON string: {"coooooool": { "name" : "coooooooooool","name" : 1"}"}n==14573== 
==14573== HEAP SUMMARY: 
==14573==  in use at exit: 0 bytes in 0 blocks 
==14573== total heap usage: 17 allocs, 17 frees, 1,511 bytes allocated 
==14573== 
==14573== All heap blocks were freed -- no leaks are possible 
==14573== 
==14573== For counts of detected and suppressed errors, rerun with: -v 
==14573== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0) 
Segmentation fault (core dumped) 
+1

为什么右括号之前的引号? –

+0

分配给'json_object * jobj'的内存在哪里? json lib是做这个还是你负责?看起来你是,但我之前没有使用过这个库,所以仔细检查。 '地址0xfffffff5不堆栈,malloc'd或(最近)free'd' ?? –

回答

0

你的问题很简单:你在你的JSON字符串错误。
您的JSON字符串是以下(不逃逸引号):

{"coooooool": { "name" : "coooooooooool","name" : 1"}"} 

两个键具有相同的名称,字符串的最后两个双引号字符是完全不合适的。
该字符串不是有效的json,因此json_tokener_parse返回NULL。
您应该执行错误检查,以赶上故障解析字符串,即增加一个检查这样的:

if(jobj == NULL) { 
    // recover from the error or quit the program 
} 

与您的代码,json_object_object_foreach收到NULL指针,引起分段错误。

+0

是的。 segfault是因为无效的json。但代码无法在json_parse()中打印值。我的意思是json_type_int。请检查我的代码 – sandesh

+0

这是正确的。这是因为代码循环包含在根对象中的所有键,并且只有一个:“coooooool”,它不是一个int,而是一个json对象。 –

+0

谢谢你的帮助 – sandesh

1

你的json无效。与格式正确的JSON字符串尝试和它的作品:

#include <stdio.h> 
#include <json/json.h> 

void json_parse(json_object * jobj); 

int main(){ 

    char * string2 = "{" 
      "\"coooooool\": { " 
      "\"name\" : \"coooooooooool\"," 
      "\"name\" : 1" 
      "\"}" 
      "\"}"; 
    char * string = "{\"name\" : \"joys of programming\"}"; 
    printf ("JSON string: %sn", string); 
    // json_object * jobj = malloc(sizeof(json_object)); 
    json_object * jobj = json_tokener_parse(string); 
    json_parse(jobj); 


    return 0; 
} 

void json_parse(json_object * jobj) { 
    enum json_type type; 
    json_object_object_foreach(jobj, key, val) { 
     type = json_object_get_type(val); 
     switch (type) 
     { 
      case json_type_int: printf("type: json_type_int, "); 
       printf("value: %dn", json_object_get_int(val)); 
       break; 
     } 
    } 
} 

您可以lint your json检查你想要的。

输出

./test1 
JSON string: {"name" : "joys of programming"} 

我编译它像这样

gcc -g -v -Wall -std=gnu99 -static -L/path/to/json-c-0.9/lib main.c -o test1 -ljson

+0

不是说无效的JSON应该会导致段错误... –

+0

是的。 segfault是因为无效的json。但代码无法在json_parse()中打印值。我的意思是json_type_int。请检查我的代码。 – sandesh