函数代码没有贴全，我估计你这个函数的其他地方还有 return 语句，但是没有返回数据，是不是直接一条 return你的方法是int型的吧，没有返回值报错了。
Im trying to connect postgresql db from my server which has ubuntu 18.04 I have the code below but when i compile it I get the error return-statement with a value, in function returning 'void' [-fpermissive] return 1; 1 2 3 4
Status Getstack(SqStack &S, SElemType e){ // 改&e 为：e, 这就允许你用常数调用。main(){ SqStack S; // 改&S 为 S if(S.top==S.base) exit(0); // 改掉 返回 return ERROR; 例如用 exit(0); 因为 void 函数体内 不能用 return 语句。50 c语言...
问返回结构时“‘return’with a value，in function return void”EN一、前言 当我们总 flutter 应用...
c++中In function 'int main()'是什么意思啊? 这种编译提示是gcc/g++的一种问题描述格式,告诉你编译问题出现在源代码什么位置。int main是指main函数需要返回一个int值。扩展资料:C++是在C语言的基础上开发的一种面向对象编程语言,应用非常广泛。常用于系统开发,引擎开
C++ - Function returning reference: Here, we will learn with a C++ program, how to return a reference from function? What is Function Returning Reference in C++? As we know that we can take only variable on the left side in C++ statements, we can also use a function on the left side...
Please enjoy;stackoverflow.com/questions/8865982/return-array-from-function-in-c constcharnumbers[] ="0123456789abcdef";voidgetBase(intn,intb,char*str) {constsize_t SIZE =32;intdigits=SIZE;while(n >0) {intt = n%b; n/=b; str[--digits] =numbers[t]; ...
function returns a pointer toconst char* and you can't just assign it to a pointer tonon-constchar (without const_cast, anyway). These are incompatible types. function1 is different, it returns a constant int. The const keyword is rather useless here - you can't change the returned value...
FunctionDescription array([expr [, …]]) Returns an array with the elements in expr. bigint(expr) Casts the value expr to BIGINT. binary(expr) Casts the value of expr to BINARY. boolean(expr) Casts expr to BOOLEAN. cast(expr AS type) Casts the value expr to the target data type ty...
Please enjoy; stackoverflow.com/questions/8865982/return-array-from-function-in-c const char numbers[] = "0123456789abcdef"; void getBase(int n, int b, char* str) { const size_t SIZE = 32; int digits=SIZE; while (n > 0) ...