The best way to declare a string literal in your code is to use array notation, like this:<\/p>\n
You can also declare a string in this manner:<\/p>\n Again, in the code you can use the Recently, I researched this phenomenon, trying to discover if I could possibly find a way around the pointer-string declaration limitation. First, I concocted a program using array notation, but manipulating the string in a function as a pointer:<\/p>\n In the change()<\/em> function, the middle character of If you change the declaration in the main()<\/em> function to read I tried everything to get the program to run, but failed. The proper address is passed to the change()<\/em> function, however any attempt to modify the string causes the program to crash with a memory violation. This situation is frustrating, especially because you can pass the string to the output()<\/em> function without a hitch.<\/p>\n The only way I can get the change()<\/em> function to modify the string pointer is when it’s not initialized. In the following code, the pointer is declared, allocated memory, then a string is assigned. After that, the program runs without a memory violation:<\/p>\n I would think that this code would run the same if I merely assigned It’s a handy shortcut, but for manipulating strings it just doesn’t work. Continue reading char string[] = \"I am some sort of interesting string.\\n\";<\/code>
\n
\nThis type of declaration is 100 percent okey-doke. The compiler allocates proper storage for the string, including the null character at the end, and it assigns it to the array string<\/code>. Nifty keen.<\/p>\nchar *string = \"I am some sort of interesting string.\\n\";<\/code><\/p>\nstring<\/code> variable to reference the text — but that’s about it. Altering the string or passing it to a function renders heaps of trouble. So, while you effectively create a string, you pretty much doom your code not to do anything with the string. In fact, there’s no clear advantage to such a declaration other than it frightens C programmers who don’t like pointers.<\/p>\n\r\n#include <stdio.h>\r\n\r\nvoid change(char *s)\r\n{\r\n *(s+1) = 'i';\r\n}\r\n\r\nvoid output(char *s)\r\n{\r\n printf(\"You are %s\\n\",s);\r\n}\r\n\r\nint main()\r\n{\r\n char string[] = \"fat\";\r\n\r\n output(string);\r\n change(string);\r\n output(string);\r\n\r\n return(0);\r\n}<\/pre>\nfat<\/code> is changed to an i<\/code>. Here’s the output:<\/p>\nYou are fat
\nYou are fit<\/code><\/p>\nchar *string = \"fat\";<\/code> the program won’t run. It compiles, but the output chokes after the first string is displayed.<\/p>\n\r\n#include <stdio.h>\r\n#include <stdlib.h>\r\n#include <string.h>\r\n\r\nvoid change(char *s)\r\n{\r\n *(s+1) = 'i';\r\n}\r\n\r\nvoid output(char *s)\r\n{\r\n printf(\"You are %s\\n\",s);\r\n}\r\n\r\nint main()\r\n{\r\n char *string;\r\n\r\n string= (char *)malloc( sizeof(char)*4 );\r\n if( string == NULL )\r\n {\r\n puts(\"Unable to allocate string storage\");\r\n exit(1);\r\n }\r\n strcpy(string,\"fat\");\r\n\r\n output(string);\r\n change(string);\r\n output(string);\r\n\r\n return(0);\r\n}<\/pre>\n*string<\/code> to a literal, but it doesn’t. And, yes, I tried passing pointers-to-pointers and other tricks. The problem is that the memory allocated by a pointer-string declaration is saved on the stack as local storage. This value isn’t available outside the function no matter how hard you try. That’s perfectly okay with me: You have plenty of alternatives for allocating string literals. And using array notation to address the issue is an easy fix.<\/p>\n","protected":false},"excerpt":{"rendered":"