C语言关于数组指针的问题（linux下的gcc环境）

Posted 2023-02-15

#include<stdio.h>
main()

int a[3][4]=0,1,2,3,4,5,6,7,8,9,10,11;
printf("%d,",a);
printf("%d,",*a);
printf("%d,",a[0]);
printf("%d,",&a[0]);
printf("%d\n",&a[0][0]);
printf("%d,",a+1);
printf("%d,",*(a+1));
printf("%d,",a[1]);
printf("%d,",&a[1]);
printf("%d\n",&a[1][0]);
printf("%d,",a+2);
printf("%d,",*(a+2));
printf("%d,",a[2]);
printf("%d,",&a[2]);
printf("%d\n",&a[2][0]);
printf("%d,",a[1]+1);
printf("%d\n",*(a+1)+1);
printf("%d,%d\n",*(a[1]+1),*(*(a+1)+1));


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以上是例题的某段程序，我在编译的时候出错了，错误信息在下边，我不明白到底有什么错

ex1.c: In function ‘main’:
ex1.c:7: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:8: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:9: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:10: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:11: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:12: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:13: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:14: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:15: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:16: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:17: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:18: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:19: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:20: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int (*)[4]’
ex1.c:21: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:22: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
ex1.c:23: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
我也知道提示是说类型不匹配，不过我看的是谭浩强的书，他说这是可以的，而且这个程序还是例题的。。。
---------------------
-1079347568,-1079347568,-1079347568,-1079347568,-1079347568
-1079347552,-1079347552,-1079347552,-1079347552,-1079347552
-1079347536,-1079347536,-1079347536,-1079347536,-1079347536
-1079347548,-1079347548
5,5
这是程序运行结果。。。和预期的完全不一样
只有最后一行没有warning提示的句子能正确输出，其他的都有问题，怎么回事？

这些都是类型不一致的警告，系统依然会生成可执行文件，在这个程序中可以忽略这些警告信息，可以用带-w参数的gcc编译这个文件，就不会输出这些警告信息了。

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你用的GCC版本是多少？在4.2.4中不会输出警告。

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如果打算在LINUX系统中学习C语言，或者打算真正掌握C语言的精华，而不是为了应付考试，就不要用这本书，这本书不是为真正学习C语言的人准备的，而是给为过考试的人设计的，里面的内容太粗糙了！而且这本书的调试环境是TC2或者VC++6.0，二者对ANSI C的支持都很差，而GCC则是对ANSI C支持最强的，所以一些在TC2或者VC++6.0中看起来正常的不标准的C程序在GCC的眼中当然会错误百出。 参考技术A 里面很多类型都不匹配~ 参考技术B 类型转换错误 跟着提示改就好了 参考技术C gcc 编译器比较严格，出现的都是警告。没事的