英文:
declaring array size shadows a local variable
问题
我在CS50第2周的数组灯泡问题中。我有一个函数将ASCII值转换为8位二进制，然后将值存储在数组中，问题是，如果不声明数组的大小，会出现错误：无法初始化可变大小的对象，如果声明数组的大小，会出现错误：声明会遮蔽局部变量。我不知道该怎么办，显然是编程新手，任何帮助都不胜感激，以下是代码：
#include <cs50.h>#include <stdio.h>#include <string.h>int main(void){char* message = get_string("message:");int i=0;int array[7]; //这是导致声明遮蔽局部变量的原因，还尝试过int array[]={0,0,0,0,0,0,0,0};//如果没有这行，下面会出现可变大小对象无法初始化的错误while (message[i] != '#include <cs50.h>#include <stdio.h>#include <string.h>int main(void){char* message = get_string("message:");int i=0;int array[7]; //这是导致声明遮蔽局部变量的原因，还尝试过int array[]={0,0,0,0,0,0,0,0};//如果没有这行，下面会出现可变大小对象无法初始化的错误while (message[i] != '\0'){int j= message[i]; //这将字符转换为ASCIIfor (int h=7; h>=0; h--){if (j % 2 == 1){array[h]=1;//如果在上面声明大小时不遮蔽这个，会出现可变大小对象无法初始化的错误j=(j/2); //我实际上不知道这是否会四舍五入到0.5，因为它是一个整数，但这真的不是我现在担心的主要问题}else{array[h]=0; //代码的目的是将消息中的每个字符打印成8位二进制j=(j/2);}for (int y=0; y < 8; y++){printf("%i",array[y]); //我明白我正在遮蔽数组，但如果我将数组的名称更改为array1，那么它就是一个新的数组，会出现可变大小对象无法初始化的错误}printf("\n");}i++;}printf("\n");}'){int j= message[i]; //这将字符转换为ASCIIfor (int h=7; h>=0; h--){if (j % 2 == 1){array[h]=1;//如果在上面声明大小时不遮蔽这个，会出现可变大小对象无法初始化的错误j=(j/2); //我实际上不知道这是否会四舍五入到0.5，因为它是一个整数，但这真的不是我现在担心的主要问题}else{array[h]=0; //代码的目的是将消息中的每个字符打印成8位二进制j=(j/2);}for (int y=0; y < 8; y++){printf("%i",array[y]); //我明白我正在遮蔽数组，但如果我将数组的名称更改为array1，那么它就是一个新的数组，会出现可变大小对象无法初始化的错误}printf("\n");}i++;}printf("\n");}
所以我确定这段代码存在很多问题，它可以更好地优化，但我真正关心的是如何在不遮蔽局部变量的情况下声明数组的大小。另外，我不得不逐行重新编写所有这些代码，从Visual Studio中复制，所以这里可能有语法错误，这还没有完成，即使修复了数组错误，我仍然需要将其打印为灯泡表情而不是1和0。
英文:
I am in CS50 week 2 arrays bulbs problem
I have a function to turn ASCII value into 8 bit binary then store value in array the problem is without declaring the size of the array it gives error: Variable-sized object may not be initialized and if I declare the size of the array it gives error: declaration shadows local variable I don't know what to do obviously new to coding any and all help is appreciated here is the code:
#include <cs50.h>#include <stdio.h>#include <string.h>int main(void){char* message = get_string("message:");int i=0;int array[7]; //this is what gives declaration shadows local variable also have tried int array[]={0,0,0,0,0,0,0,0};//without this line I get the variable-sized object may not be initialized down bellowwhile (message[i] != '#include <cs50.h>#include <stdio.h>#include <string.h>int main(void){char* message = get_string("message:");int i=0;int array[7]; //this is what gives declaration shadows local variable also have tried int array[]={0,0,0,0,0,0,0,0};//without this line I get the variable-sized object may not be initialized down bellowwhile (message[i] != '\0'){int j= message[i]; // this converts char to ASCIIfor ( h=7;h>=0;h--){if (j % 2 == 1){int array[h]=1;// if I don't shadow this when declaring size above I get variable-sized// object may not be initializedj=(j/2); // I don't actually know if this will result rounding down the .5 since its an//int but that's really the least of my worries right now}else{int array[h]=0; //the point of the code is to print each of the chars in the message to an 8 digit binaryj=(j/2);}for ( int y=0;y < 8; y++){printf("%i",array[y]); //I understand that I am shadowing the array but if i change the name of the array to array1 then its a new array and would get error variable sized object may not be initialized}printf("\n");i++;}}printf("\n");}'){int j= message[i]; // this converts char to ASCIIfor ( h=7;h>=0;h--){if (j % 2 == 1){int array[h]=1;// if I don't shadow this when declaring size above I get variable-sized// object may not be initializedj=(j/2); // I don't actually know if this will result rounding down the .5 since its an//int but that's really the least of my worries right now}else{int array[h]=0; //the point of the code is to print each of the chars in the message to an 8 digit binaryj=(j/2);}for ( int y=0;y < 8; y++){printf("%i",array[y]); //I understand that I am shadowing the array but if i change the name of the array to array1 then its a new array and would get error variable sized object may not be initialized}printf("\n");i++;}}printf("\n");}
so I'm sure there are plenty of problems with this code and that it could be better optimized but all I really care about is how I can declare the size of the array without shadowing local variable. Also I had to rewrite all this code line by line copying what I have in visual studio so there could be an syntax error somewhere here that's not in my actual code this is unfinished even with the array error fixed I will still need to print it as light emoji and not ones and zeros.
答案1
得分: 2
这是一个int类型的数组声明（大小为7）
int array[7];
这与上面的声明相同，但带有初始化
int array[7] = {0, 1, 2, 3, 4, 5, 6};
进一步的代码可以通过标识符array访问该数组，如下所示
array[h] = 42;
但如果你在上面的表达式前面加上关键字int（例如int array[h] = {42};），那是一个新的int类型数组声明（大小为h）。由于已经存在具有该名称的变量，新声明会“遮蔽”以前的声明。因此编译器会发出警告（以确保你知道自己在做什么）。
英文:
This is a declaration of an array of type int (of size 7)
int array[7];
this is the same declaration as above, but with initialization
int array[7] = {0, 1, 2, 3, 4, 5, 6};
Further code can access that array via the identifier array, like this
array[h] = 42;
but if you prefix the above expression with the keyword int (e.g. int array[h] = {42};), that is a new declaration of an array of type int (of size h). Since there already exists a variable with that name, the new declaration 'shadows' the previous one. Therefore the warning from the compiler (to make sure, that you know what you're doing).
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