c-strings

问题 Possible Duplicate: What is the difference between char s[] and char *s in C? Question about pointers and strings in C I'm reading about the strings in C and I'm confused. I can "declare" strings in two ways: char *str = "This is string"; char str2[20] = "This is string"; What is the difference between the two declarations? When would char str2[20] be preferred over char *str ? 回答1: char *str = "This is string"; Puts the string in the constant data section (also known as .rdata) of the

问题 This is from a small library that I found online: const char* GetHandStateBrief(const PostFlopState* state) { static std::ostringstream out; // ... rest of the function ... return out.str().c_str() } In my code I am doing this: const char *d = GetHandStateBrief(&post); std::cout<< d << std::endl; Now, at first d contained garbage. I then realized that the C string I am getting from the function is destroyed when the function returns because std::ostringstream is allocated on the stack. So I

问题 I saw use of this pattern to concatenate onto a string in some code I was working on: sprintf(buffer, "%s <input type='file' name='%s' />\r\n", buffer, id); sprintf(buffer, "%s</td>", buffer); and I'm fairly certain it's not safe C. You'll notice that buffer is both the output and the first input. Apart from the obvious possibility of a buffer overflow , I believe there is no guarantee that buffer doesn't get changed between the start and the end of the function (i.e., there is no guarantee

问题 This question already has an answer here: Why is the gets function so dangerous that it should not be used? 11 answers While using gets() in my code, the compiler shouts warning: the 'gets' function is dangerous and should not be used.` and warning: ‘gets’ is deprecated (declared at /usr/include/stdio.h:638) [-Wdeprecated-declarations] Any specific reasons? 回答1: Can someone explains why the compiler shows like that…? Yes, because, the gets() function is dangerous, as it suffers from buffer

问题 I need a help on one question where I stuck while coding my app in MFC . I am using CLR i.e Common Language Runtime in my application to integrate c# APIs. but now I stuck on converting System::String^ to CString . I am not able to do that. I am using Following code. String^ csPass = gcnew String(strPassword.GetBuffer()); array<Byte>^ Value = Encoding::UTF8->GetBytes(csPass); for (int i = 0; i < Value->Length; i++ ) { csPass += String::Format( "{0:X2}", Value[ i ] ); } now I want to convert

问题 When terminating a string, it seems to me that logically char c=0 is equivalent to char c=\'\\0\' , since the \"null\" (ASCII 0) byte is 0 , but usually people tend to do \'\\0\' instead. Is this purely out of preference or should it be a better \"practice\"? What is the preferred choice? EDIT: K&R says : \"The character constant \'\\0\' represents the character with value zero, the null character. \'\\0\' is often written instead of 0 to emphasize the character nature of some expression, but

问题 I have this code: char *name = \"George\" if(name == \"George\") printf(\"It\'s George\") I thought that c strings could not be compared with == sign and I have to use strcmp . For unknown reason when I compile with gcc (version 4.7.3) this code works. I though that this was wrong because it is like comparing pointers so I searched in google and many people say that it\'s wrong and comparing with == can\'t be done. So why this comparing method works ? 回答1: I thought that c strings could not

问题 I\'m not completely sure how to do this in C: char* curToken = strtok(string, \";\"); //curToken = \"ls -l\" we will say //I need a array of strings containing \"ls\", \"-l\", and NULL for execvp() How would I go about doing this? 回答1: Since you've already looked into strtok just continue down the same path and split your string using space ( ' ' ) as a delimiter, then use something as realloc to increase the size of the array containing the elements to be passed to execvp . See the below

问题 The following code receives seg fault on line 2: char *str = \"string\"; str[0] = \'z\'; // could be also written as *str = \'z\' printf(\"%s\\n\", str); While this works perfectly well: char str[] = \"string\"; str[0] = \'z\'; printf(\"%s\\n\", str); Tested with MSVC and GCC. 回答1: See the C FAQ, Question 1.32 Q : What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i] . A : A