内存分配

发布于 2024-11-10 12:13:04 字数 144 浏览 2 评论 0原文

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 
Values val[16];

如何为结构体中的数字分配内存？

原文

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 
Values val[16];

How can one allocate memory for numbers in the struct?

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酷到爆炸 2024-11-17 12:13:05

如果您想在结构体的第一个元素中为一个浮点数分配空间，您可以这样做：

#include <stdlib.h>

/* ... */

val[0].numbers = malloc(1 * sizeof(float)); /* "1 * " for clarity... */

如果这就是您的意思。

If you want to allocate the space for one float in the first element of your struct, you would do it like this:

#include <stdlib.h>

/* ... */

val[0].numbers = malloc(1 * sizeof(float)); /* "1 * " for clarity... */

If that's what you meant.

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任谁 2024-11-17 12:13:05

因此，您可以在不使用动态内存分配的情况下执行此操作，如下所示：

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    float myfloats[MAX_SIZE];
    int i;
    for(i=0;i<MAX_SIZE;i++) {
        val[i].numbers=&myfloats[i];
    }
    return 0;
 }

但我想不出您想要一个仅具有指向一个浮点的指针的结构的任何原因。

根据名称 'numbers'，我想说您希望 'numbers' 指向 floats 数组，如果是这样，您可以这样做：

#include <malloc.h>

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    size_t numberOfFloats = 10;
    int i;

    // for each of the members of the val array
    for(i=0;i<MAX_SIZE;i++) {
        // allocate using calloc (this will set all of the floats to 0.0)
        val[i].numbers=calloc(numberOfFloats,sizeof(float));

        // check the allocation worked...
        if(!val[i].numbers) {
            // insert proper error handling here.
            printf("oops\n");
            return -1;
        }

    }

    // you access the variables like this
    for(i=0;i<MAX_SIZE;i++) {
        int number;
        for(number=0; number < numberOfFloats; number++) {
            printf("Value %d, Number %d = %f\n",i,number,val[i].numbers[number]);
        }
    }

    // don't forget to play nice and clean up afterwards
    for(i=0;i<MAX_SIZE;i++) {
        free(val[i].numbers);
    }
    return 0;
 }

So, you could do this without using dynamic memory allocation like this:

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    float myfloats[MAX_SIZE];
    int i;
    for(i=0;i<MAX_SIZE;i++) {
        val[i].numbers=&myfloats[i];
    }
    return 0;
 }

But I can't think of any reason why you'd want a structure with a pointer to just one float.

Based on the name 'numbers', I'd say you want 'numbers' to point to an array of floats, if so, you could do this:

#include <malloc.h>

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    size_t numberOfFloats = 10;
    int i;

    // for each of the members of the val array
    for(i=0;i<MAX_SIZE;i++) {
        // allocate using calloc (this will set all of the floats to 0.0)
        val[i].numbers=calloc(numberOfFloats,sizeof(float));

        // check the allocation worked...
        if(!val[i].numbers) {
            // insert proper error handling here.
            printf("oops\n");
            return -1;
        }

    }

    // you access the variables like this
    for(i=0;i<MAX_SIZE;i++) {
        int number;
        for(number=0; number < numberOfFloats; number++) {
            printf("Value %d, Number %d = %f\n",i,number,val[i].numbers[number]);
        }
    }

    // don't forget to play nice and clean up afterwards
    for(i=0;i<MAX_SIZE;i++) {
        free(val[i].numbers);
    }
    return 0;
 }

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