C指针，指向混乱 [英] C Pointer confusion

问题描述

我要存储在内存中的字符串，后来阅读：

  $$  - ＆GT; desc.constant-＆GT; base.id =（字符*）malloc的（200）;
sprintf的（$$  - ＆GT; desc.constant-＆GT; base.id，％F，$ 1）;
的printf（ - ＆GT;％S \\ n，$$  - ＆GT; desc.constant-＆GT; base.id）; // A线
的printf（ - ＆GT;％I \\ N，$$  - ＆GT; desc.constant）; // B线//其他一些code//然后，后来在一个函数调用：的printf（％i的，expr-＆GT; desc.constant）; // D线
的printf（％S，expr-＆GT; desc.constant-＆GT; base.id）; // C线
 

虽然B线和D线显示了相同的地址，在C线的printf的失败，分段故障。我缺少什么？

任何帮助真的会AP preciated！

解决方案

 的printf（ - ＆GT;％I \\ N，$$  - ＆GT; desc.constant）; // B线
 

这是无效的。当你表现出线条在它之前的恒实际上是一个指针，你不能把它当作好像它是类型 INT 。它们不necassarily具有相同的sizeof和对准。使用用于无效* 的格式。它会输出内存地址正确：

 的printf（ - ＆GT;％P \\ N（无效*）$$  - ＆GT; desc.constant）; // B线
 

I want to store a string in memory and read it later:

$$->desc.constant->base.id =  (char*)malloc(200);
sprintf($$->desc.constant->base.id, "%f", $1);
printf("->%s\n", $$->desc.constant->base.id); //LINE A
printf("->%i\n", $$->desc.constant); //LINE B

//SOME OTHER CODE

//Then, later on in a function call:

printf("%i", expr->desc.constant); // LINE D
printf("%s", expr->desc.constant->base.id); // LINE C

Although Line B and Line D show the same address, the printf in Line C fails with a Segmentation fault. What am I missing?

Any help would really be appreciated!

解决方案

printf("->%i\n", $$->desc.constant); //LINE B

That is invalid. As you show the line prior to it that constant is actually a pointer, you cannot treat it as if it were of type int. They don't necassarily have the same sizeof and alignment. Use the format used for void*. It will output memory addresses properly:

printf("->%p\n", (void*)$$->desc.constant); //LINE B

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