C语言 c 中对函数错误的未定义引用
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undefined reference to function error in c
提问by heidi boynww
#include <stdio.h>
int singleFib(int x,int a,int b);
int multiFib(int x);
void main(){
int n,i;
printf("How much?\n");
scanf("%d",&n);
for(i=0;i<n;i++){
printf("%5d. number: %d - %d \n",i+1,multiFib(i),singleFib(i,1,1));
}
getch();
return 0;
}
int multiFib(int x){
if (x<2){
return 1;
}
else{
return multiFib(x-2)+multiFib(x-1);
}
int singleFib(int x,int a,int b){
if (x<2){
return 1;
}
else{
return singleFib( x-1, b,(a+b));
}
}
}
The error is in
错误在
singleFib(i,1,1) in `printf`
Whyis that problem?how can i solve that problem? i am using codeblocks
为什么是这个问题？我该如何解决这个问题？我正在使用代码块
Codeblocks\Fiberonacci\main.c|14|undefined reference to `singleFib'| The error is that.how can i solve?
Codeblocks\Fiberonacci\main.c|14|对`singleFib'的未定义引用| 错误是那个。我该如何解决？
回答by godel9
You're missing a close bracket
}at the end of yourmultiFibfunction.You have an extra close bracket
}at the end of yoursingleFibfunction.The function
mainshould have return typeint.
您在函数
}末尾缺少一个右括号multiFib。}在singleFib函数的末尾有一个额外的右括号。函数
main应该有返回类型int。
回答by heidi boynww
The {and }in multiFiband singleFibare mixed up, singleFibis declared inside multiFib:
该{和}在multiFib和singleFib混合起来，singleFib声明内multiFib：
int multiFib(int x){
if (x<2){
return 1;
}
else{
return multiFib(x-2)+multiFib(x-1);
}
/*************************************/
int singleFib(int x,int a,int b) {
if (x<2){
return 1;
}
else{
return singleFib( x-1, b,(a+b));
}
}
/*************************************/
}
it works in gcc, since it's an nonstandard nested functionsextension, but the function will not be accessible outside multiFib.
它适用于 gcc，因为它是一个非标准的嵌套函数扩展，但该函数在multiFib.
回答by aks
Seems like there is a mismatch in the braces of the function definition
函数定义的大括号似乎不匹配
回答by Eutherpy
You're missing a closing bracket for multiFib, but have one that you don't need after singleFib.
您缺少 的右括号multiFib，但有一个您在 之后不需要的括号singleFib。
