[問題] 指標陣列算size大小
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遇到的問題： (題意請描述清楚)
怎麼算出char *keytab[]的大小
題目原文:
char *keytab[]={"break","case","char","continue","default","for","if","int"};
char **p=keytab;
please use sizeof operator to decide the number of element of keytab
array and define it as NKEYS.
希望得到的正確結果：
36 因為我想說 break case char continue default for if int
這些字元一共有36個
所以我猜答案是36
程式跑出來的錯誤結果：
32
開發平台： (例: VC++ or gcc/g++ or Dev-C++, Windows or Linux)
Dev-C++ Windows
有問題的code： (請善用置底文標色功能)
#include <iostream>//III
using namespace std;
int main()
{
char *keytab[]={"break","case","char","continue","default","for","if","int"};
char **p=keytab;
for (int i=0; i<8 ;i++)
{
cout<<keytab[i]<<endl;
}
int NKEYS;
NKEYS = sizeof(keytab)/sizeof(char);
cout<<NKEYS;
cout<<sizeof(keytab); //32
system("pause");
return 0;
}
補充說明：
我知道 sizeof(keytab); 是算出8個指標 8X4=32的意思
請問要怎麼用sizeof 去算出 一共使用了36個字元呢????
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※ 編輯: nana0130 來自: 140.116.39.115 (05/03 13:38)
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