在函数指针代码中显示错误
我写了一个使用函数指针来比较字符串的代码。但是,它显示了我的错误,我不知道如何纠正它们。下面是代码:在函数指针代码中显示错误
#include<stdio.h>
#include<string.h>
void sports_no_bieber(char *);
void science_sports(char *);
void theater_no_guys(char *);
int find(int(*match)(char*));
int NUM_ADS=4;
char *ADS[]={
"Sarah:girls, sports, science",
"William: sports, TV, dining",
"Matt: art, movies, theater",
"Luis: books, theater, guys",
"Josh: sports, movies, theater"
};
int main()
{
printf("Bachelorette Amanda needs your help! He wants someone who likes sports but not bieber.\n");
find(sports_no_bieber);
printf("Bachelorette Susan needs your help! She wants someone who likes science and sports. (And girls).\n");
find(science_sports);
printf("Bachelorette Emily needs your help! She wants someone who likes theater but not guys.\n");
find(theater_no_guys);
return 0;
}
int find(int(*match)(char*))
{
int i;
puts("Search results\n");
puts("--------------------");
for(i=0;i<NUM_ADS;i++)
{
if(match(ADS[i]))
printf("%s\n",ADS[i];
}
puts("--------------------");
return i;
}
int sports_no_bieber(char * s)
{
return (strstr(s, "sports")) && (!strstr (s,"bieber"));
}
int science_sports(char * s)
{
return (strstr(s, "science")) && (strstr (s,"sports"));
}
int theater_no_guys(char * s)
{
return (strstr(s, "theater"))&&(!strstr(s,"guys"));
}
,并显示错误是
E:\ComputerPrograming\FunctionPointer.c: In function `int main()':
E:\ComputerPrograming\FunctionPointer.c:18: passing `void (*)(char *)' as argument 1 of `find(int (*)(char *))'
E:\ComputerPrograming\FunctionPointer.c:20: passing `void (*)(char *)' as argument 1 of `find(int (*)(char *))'
E:\ComputerPrograming\FunctionPointer.c:22: passing `void (*)(char *)' as argument 1 of `find(int (*)(char *))'
E:\ComputerPrograming\FunctionPointer.c: In function `int find(int (*)(char *))':
E:\ComputerPrograming\FunctionPointer.c:36: parse error before `;'
E:\ComputerPrograming\FunctionPointer.c:40: confused by earlier errors, bailing out
我甚至试图使查找功能为一个int功能......但没有任何区别。错误究竟意味着什么?
这些函数声明:
void sports_no_bieber(char *);
void science_sports(char *);
void theater_no_guys(char *);
不要find()
或它们的定义所需的函数指针的签名相匹配。更改为:
int sports_no_bieber(char *);
int science_sports(char *);
int theater_no_guys(char *);
注意NUM_ADS
不等于ADS
数组中元素的数目:它是一个更小。为了避免必须确保NUM_ADS
和ADS
与NULL
指针被正确终止NUM_ADS
阵列,并用它作为循环终止条件(并丢弃NUM_ADS
):
const char *ADS[] =
{
"Sarah:girls, sports, science",
"William: sports, TV, dining",
"Matt: art, movies, theater",
"Luis: books, theater, guys",
"Josh: sports, movies, theater",
NULL
};
for(int i=0; ADS[i]; i++)
{
推荐使所有功能的参数类型为const char* const
代替因为没有任何功能修改内容或重新分配指针。
啊。感谢您指出了这一点。这对我来说相当愚蠢。 – 2013-05-08 12:45:46
您有两种类型的错误,首先是原型和函数本身之间的不匹配。
void sports_no_bieber(char *);
^ ^ ^
| | |
these much mach the types here must
exactly match
| | |
v v v
int sports_no_bieber(char * s)
因此,您需要名称和返回类型相同,就像使用参数类型一样。在您的情况下,退货类型不匹配sports_no_bieber()
,science_sports()
和theater_no_guys()
。
避免这个问题的一种方法是将功能定义移到使用点上方,这样就不需要原型,并消除了误弄它们的机会......当然,您也可以复制和粘贴以避免这样的愚蠢错误。
您有其他错误是在你的find()
功能,你错过了一个括号:
printf("%s\n",ADS[i]; // <-- missed the close)
你的函数原型不匹配你的函数定义... – Mat 2013-05-08 12:36:55