Question

以下是我可以创建的问题的最短的可编辑演示。

查看结尾附近的printf电话，输出如下。我不知道为什么完全相同的两个语句可能会这样做。它可能与输入类型有关，但我看不出它可能是什么。

#include <gtk/gtk.h>
#include <libxml/xmlmemory.h>
#include <libxml/parser.h>
int
main (int argc, char *argv[])
{
    gtk_init (&argc, &argv);
    GtkBuilder * builder = gtk_builder_new ();
    GHashTable * table = g_hash_table_new(NULL,NULL);
    int i;
    char builderdoc[] = "<?xml version=\"1.0\"?>"
"<interface>"
"  <requires lib=\"gtk+\" version=\"2.16\"/>"
 " <!-- interface-naming-policy project-wide -->"
"  <object class=\"GtkCheckButton\" id=\"checkbutton1\">"
"  </object>"
"</interface>";

    // Load gtkbuilder
    gtk_builder_add_from_string (builder, builderdoc, sizeof(builderdoc),NULL);

    char * buffer[][2] = {
    {"log","checkbutton1"}
    };

    // Load array of widgets into hash table (Shortened)
    for(i = 0; i < sizeof(buffer) / sizeof(char *) / 2; i++){
        g_hash_table_insert(table,(gchar *) buffer[i][0],gtk_builder_get_object (builder, (gchar *) buffer[i][1]));
    }


    // Load xml doc
    xmlDocPtr doc;
    xmlNodePtr cur;
    xmlNodePtr cur2;
    char xmldoc[] = "<?xml version=\"1.0\"?>"
                    "<CsSettings>"
                    "<options>"
                    "<check name=\"log\" value=\"1\" />"
                    "</options>"
                    "</CsSettings>";

    doc = xmlParseMemory(xmldoc, (int) sizeof(xmldoc));
    cur = xmlDocGetRootElement(doc);
    cur = cur->xmlChildrenNode;

    // Find what we're looking for
    while (cur != NULL){
        if(xmlStrEqual(cur->name, (xmlChar *) "options")){
            cur2 = cur->xmlChildrenNode;
            while (cur2 != NULL){
                if(xmlStrEqual(cur2->name, (xmlChar *) "check")){
                    // We've found it, now print some output
                    printf("Plain old lookup: g_hash_table_lookup(table,\"log\"): %p\n",g_hash_table_lookup(table,"log"));
                    printf("Variable lookup : g_hash_table_lookup(table,\"%s\"): %p\n",(gchar *) xmlGetProp(cur2, (xmlChar *) "name"),g_hash_table_lookup(table,(gchar *) xmlGetProp(cur2, (xmlChar *) "name")));
                    printf("Xml says we should lookup: '%s'\n",(gchar *) xmlGetProp(cur2, (xmlChar *) "name"));

                }
                cur2 = cur2->next;
            }
        }       
        cur = cur->next;
    }

}

令人眼花缭乱的输出：

Plain old lookup: g_hash_table_lookup(table,"log"): 0x1e369a0
Variable lookup : g_hash_table_lookup(table,"log"): (nil)
Xml says we should lookup: 'log'

Answer 1

您可能会从哈希表中获得意外行为，因为您没有在构造函数上设置任何哈希或相等的函数，在这种情况下，对g_direct_hash()和g_direct_equal()函数的指针进行直接哈希处理

如果表格的密钥为gchar *，我会使用g_str_hash和g_str_equal。这样的东西可以用来创建你的桌子：

GHashTable * table = g_hash_table_new(g_str_hash,g_str_equal);

如果哈希表的键是指针指向的内容，则指针上的直接哈希不是一个好主意。请告诉我们这是否解决了这个问题。

GHashTable对两个相同的输入完全不同

1 个答案: