如果要這樣做，則需要傳入指向指針的指針。void?barPush(BarList?**?list,Bar?*?bar){

if?(list?==?NULL)?return;?//?need?to?pass?in?the?pointer?to?your?pointer?to?your?list.

//?if?there?is?no?move?to?add,?then?we?are?done

if?(bar?==?NULL)?return;

//?allocate?space?for?the?new?node

BarList?*?newNode?=?malloc(sizeof(BarList));

//?assign?the?right?values

newNode->val?=?bar;

newNode->nextBar?=?*list;

//?and?set?the?contents?of?the?pointer?to?the?pointer?to?the?head?of?the?list

//?(ie:?the?pointer?the?the?head?of?the?list)?to?the?new?node.

*list?=?newNode;?}

然后像這樣使用它：BarList?*?l;l?=?EMPTY_LIST;barPush(&l,&b1);?//?b1?and?b2?are?just?Bar'sbarPush(&l,&b2);

喬納森·萊弗勒(JonathanLeffler)建議將評論中的新頭目返回：BarList?*barPush(BarList?*list,Bar?*bar){

//?if?there?is?no?move?to?add,?then?we?are?done?-?return?unmodified?list.

if?(bar?==?NULL)?return?list;

//?allocate?space?for?the?new?node

BarList?*?newNode?=?malloc(sizeof(BarList));

//?assign?the?right?values

newNode->val?=?bar;

newNode->nextBar?=?list;

//?return?the?new?head?of?the?list.

return?newNode;?}

用法如下：BarList?*?l;l?=?EMPTY_LIST;l?=?barPush(l,&b1);?//?b1?and?b2?are?just?Bar'sl?=?barPush(l,&b2);

總結

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