类型为“char（*）[16]”的参数与“const char *”类型的参数不兼容 [英] Argument of type "char (*)[16]" is incompatible with parameter of type "const char *"

问题描述

你好，

我正在使用visual studio 2015执行交流程序

i我试图从文件中读取数据，但我收到错误：'文件* fopen（const char *，const char *）'：不能将参数1从'char（*）[16]'转换为'const char *'

以及

类型char（*）[16]的参数与const char *类型的参数不兼容

有人可以帮我解决这个问题

感谢你，

spanika kamuni



我尝试过：

hello,
I am using visual studio 2015 to perform a c program
i am trying to read the data from a file, but i am getting error:'FILE *fopen(const char *,const char *)': cannot convert argument 1 from 'char (*)[16]' to 'const char *'
and also
argument of type "char (*)[16]" is incompatible with parameter of type "const char *"
can someone please help me to fix this
Thanking you,
spanika kamuni

What I have tried:

include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <conio.h>
#include <iostream>
#include <string>
#include "stdafx.h" 

#define MAXNAME 16

void main(int argc, char *argv[])
{
	int c, i, j, f;
	char lib[MAXNAME] = "lib\0       ";
	FILE *in, *out;
	
	char name[MAXNAME];
	
	/* open input file */

	i = 0;
	while ((name[i] = *argv[1]) != '\0') {
		i++;
		argv[1]++;
	}
	if ((in = fopen(&name, "r")) == NULL) {
		printf("Can't open file\n");
		goto end;
	}

	/* open output file. Same name as input but with .lib extension */

	while (name[i] != '.') i--;
	j = 0;
	i++;
	while (MAXNAME >= (i + j)) {
		name[i + j] = lib[j];
		j++;
	}
	if ((out = fopen(&name, "w")) == NULL) {
		printf("Can't open file\n");
		goto end;
	}
}

推荐答案

您的变量 name 是一个数组，因此您不需要在其前面使用addressof运算符（&）。它应该是：

Your variable name is an array, so you do not need the addressof operator (&) in front of it. It should be:

if ((in = fopen(name, "r")) == NULL) {

和输出文件相同。

and the same for the output file.

这篇关于类型为“char（*）[16]”的参数与“const char *”类型的参数不兼容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！