char * ptr =" amit"，存储它的内存区域（堆，堆栈，代码，全局） [英] char * ptr = "amit", in which area of memory it is stored (heap,stack,code,global)

问题描述

char * str1 =" amit";

char str2 [] =" mehta"

strcat（str1，str2）;


它会崩溃，我觉得str1会存储在代码部分。一旦分配了内存

，你就无法更改或附加到此内容。


请纠正我如果我错了..

解决方案

Amit写道：

char * str1 =" amit" ;;
char str2 [] = mehta
strcat（str1，str2）;

它会崩溃，我觉得str1会存储在代码段中。一旦分配了内存，就无法更改或附加到此内容。

请纠正我如果我错了..

你很接近。

C的规则说

试图写一个由字符串文字标识的对象，

导致未定义的行为。

str1指向由字符串文字标识的对象。

此外，您需要确保有足够的空间来进行strcat。


char * str1 =" amit";

char str2 [sizeof" amit" - 1 + sizeof" mehta"] =" mehta";


strcat（str2，str1）;


-

pete

Amit写道：

char * str1 =" amit" ;;
char str2 [] =" mehta"
strcat（str1，str2）;

它会崩溃，我觉得str1会存储在代码段中。一旦分配了内存，你就无法改变或追加到它。

请纠正我如果我错了..

我觉得它是错了，因为

str1的地址值为amit

u必须strcat（* str1，str2）

venkatesh写道：

Amit写道：

char * str1 =" amit" ;;
char str2 [] =" mehta"
strcat（str1，str2）;

它会崩溃，我觉得str1会存储在代码部分。
一旦分配了内存，你就不能改变或追加到这个。

请纠正我如果我错了..

我认为这是错的，因为
str1的地址是值amit
你必须strcat（* str1，str2）

让你自己去编译器并尝试你的想法。

你错了。


* str1是char类型的表达式，其值为'a''，

引用不可写对象中的一个字节。

strcat的第一个参数必须是指向字符串的指针

在一个可写对象中，有足够的空间来生成

连接字符串。


-

pete

char *str1="amit";
char str2[]="mehta"
strcat(str1,str2);

It will crash, I feel str1 will be stored in code section. Once memory
is allocated, you cannot change or append into this.

Please correct me If I am wrong..

解决方案

Amit wrote:

char *str1="amit";
char str2[]="mehta"
strcat(str1,str2);

It will crash, I feel str1 will be stored in code section. Once memory
is allocated, you cannot change or append into this.

Please correct me If I am wrong..

You''re close.
The rules of C say that
an attempt to write to an object identified by a string literal,
causes undefined behavior.
str1 points to an object identified by a string literal.
Also, you need to make sure you have enough room for strcat.

char *str1 = "amit";
char str2[sizeof "amit" - 1 + sizeof "mehta"] = "mehta";

strcat(str2, str1);

--
pete

Amit wrote:

char *str1="amit";
char str2[]="mehta"
strcat(str1,str2);

It will crash, I feel str1 will be stored in code section. Once memory
is allocated, you cannot change or append into this.

Please correct me If I am wrong..

I thing it is wrong because
str1 has the address of the value amit
u have to strcat(*str1,str2)

venkatesh wrote:

Amit wrote:

char *str1="amit";
char str2[]="mehta"
strcat(str1,str2);

It will crash, I feel str1 will be stored in code section.
Once memory
is allocated, you cannot change or append into this.

Please correct me If I am wrong..

I thing it is wrong because
str1 has the address of the value amit
u have to strcat(*str1,str2)

Get yourself to a compiler and try your idea.
You''re very wrong.

*str1 is an expression of type char with a value of ''a'',
refering to a byte in a nonwriteable object.
The first argument to strcat must be a pointer to a string
in a writable object that has enough space for the resulting
concatenated string.

--
pete

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