当数据为NULL时访问链接列表的数据元素时应用程序崩溃
当数据为NULL时访问链接列表的数据元素时应用程序崩溃
提问于 2019-05-23 11:19:12
我使用以下代码遍历目标c中的链接列表
const MSList *calls = linphone_core_get_calls(LC);
if (calls == NULL)
{
[self dismissCtrl];
//how to check current is which screen is on
// while ((currentView == CallView.compositeViewDescription) ||
// (currentView == CallIncomingView.compositeViewDescription) ||
// (currentView == CallOutgoingView.compositeViewDescription)) {
// [self popCurrentView];
// }
} else {
linphone_call_resume((LinphoneCall *)calls->data);
while (calls)
{
if(calls->data != NULL && calls->data != nil && calls->data != (__bridge void *)((id)[NSNull null]))
{
//crash
if (linphone_call_get_state((LinphoneCall *)calls->data) == LinphoneCallIncomingReceived ||
linphone_call_get_state((LinphoneCall *)calls->data) == LinphoneCallIncomingEarlyMedia) {
[self displayIncomingCall:(LinphoneCall *)calls->data];
break;
}
}
calls = calls->next;
}当整个列表不是空，但它的数据、前一个或下一个值为空时，应用程序就会崩溃。我添加了代码来检查数据是否为NULL，但如果数据(数据)为NULL，那么我将无法访问它，并且在其本身处于崩溃状态的情况下，应用程序将崩溃。如何防止这种情况发生？我已经为应用程序崩溃的位置和列表当时包含的值附加了屏幕快照。
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-07-11 06:08:21
if (calls == NULL) {
[self dismissCtrl];
//how to check current is which screen is on
// while ((currentView == CallView.compositeViewDescription) ||
// (currentView == CallIncomingView.compositeViewDescription) ||
// (currentView == CallOutgoingView.compositeViewDescription)) {
// [self popCurrentView];
// }
}
else
{
size_t count = bctbx_list_size(calls);
linphone_call_resume((LinphoneCall *)calls->data);
int i = 0;
while (calls )
{
if(i < count)
{
if ( calls->data == (__bridge void *)((id)[NSNull null]) || calls->data == NULL || calls->data == nil )
{
return;
}
LinphoneCall *objCall = (LinphoneCall *)calls->data;
LinphoneCallState state = (objCall != NULL) ? linphone_call_get_state(call) : 0;
if (state == LinphoneCallIncomingReceived ||
state == LinphoneCallIncomingEarlyMedia)
{
[self displayIncomingCall:(LinphoneCall *)calls->data];
break;
}
//calls ? calls->data : NULL
// calls = calls ? calls->next : NULL;
calls = calls->next;
i = i+1;
}
else
{
break;
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56274086
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